^^ .<^* #^^ QforttEll IHmueraitg Eihrarij atljata, Nem §ark ..Xo.h.n M.e.nr.y..j[a.jnjn-er:.. Cornell University Library QA 372.P55 Differential equations, 3 1924 001 549 850 Cornell University Library The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001549850 WORKS OF H. B. PHILLIPS, PH.D. PUBLISHED BY JOHN WILEY & SONS, Inc. Differential Equations. V + 76 pages. 5 by 7Ji- Illustrated. Cloth, $1.25 net. Analytic Geometry. vii+197 pages. 5 by 7M. Illustrated. Cloth, $1.76 net. Differential Calculus. V + 162 pages. 5 by 1)4,. Illustrated. Cloth, $1.50 net. Integral Calculus. V + 194 $1.75 nei V + 194 pages. 5 by 1%. Illustrated. Cloth, t. Differential and Integral Calculus. In one volume. $2.50 net. DIFFERENTIAL EQUATIONS BY H. B. PHILLIPS, Ph. D. Associate Professor of Mathematics in the Massachusetts Institute of Technology NEW YORK JOHN WILEY & SONS, Inc. London: CHAPMAN & HALL, Limited 1922 4i^o:& COPTBIGHT, 1922 BY H. B. PHILLIPS TECHNICAL COMPOSITION CO. CAMBBIDQE!, MASS., XT. ». A. PREFACE With the formal exercise in solving the types of ordinary differential equations that usually occur it is the object of this text to combine a thorough drill in the solution of prob- lems in which the student sets up and integrates his own differential equation. For this purpose certain topics in mechanics and physics needed in groups of problems are briefly presented in the text. The problems have been collected from a variety of sources among which the author wishes particularly to mention the Advanced Calculus of Professor E. B. Wilson and the notes on Mathematics for Chemists prepared by Professors W. K. Lewis and F. L. Hitchcock. H. B. PHILLIPS Cambridge, Mass., Feb. 15, 19^2. CONTENTS CHAPTER I AST. Variables Separable page 1. Definitions 1 2. Separation of the variables 1 3. Different forms of solution 2 4. Derivative relations 4 5. Determination of !!he constants 5 6. Differential relations 7 7. Flow of water from an orifice 9 S. Equation of continuity 10 9. Flow of heat 11 10. Second order processes 12 CHAPTER II Other First Order Equations V il. Exact differential equations 19 12. Integrating factors 21 13. Linear equations 21 14. Equations reducible to linear form 23 15. Homogeneous equations 25 16. Change of variable 26 17. Simultaneous equations 28 CHAPTER III Special Types of Second Order Equations 18. Equations immediately integrable 33 19. Equations not containing y 33 20. Equations not containing x 34 21. Deflection of beams 37 22. Equilibrium of a cable 39 23. Motion of a particle in a straight line 40 V VI CONTENTS ART. PAGE 24. Motion of the center of gravity 42 25. Motion in a plane 43 26. Rotation about an axis 47 27. Combined translation and rotation . . 48 CHAPTER IV LiNBAK Equations with Constant Coefficients 28. Equations of the n-th order 55 29. Linear equations with constant coefficients 56 30. Equations with right hand member zero 57 31. Equations with right hand member a function of a; 60 32. Simultaneous equations 64 33. Vibrating systems 66 DIFFERENTIAL EQUATIONS CHAPTER I FIRST ORDER EQUATIONS, VARIABLES SEPARABLE 1. Definitions. — In this chapter we consider problems in- volving two variables one of which is a function of the other. It is often possible from the statement of a problem to ob- tain an equation involving the differentials or the derivatives of the variables. Such an equation is called a differential equation. Thus {j? + y^) dx + 2xydy = and dx^ dx are differential equations. A solution of a differential equation is an equation con- necting the variables such that if the derivatives are cal- culated from it and substituted in the differential equation, the latter will be satisfied. Thus y = x^ — 2x is a solution of the second equation above; for when x^ — 2 x is substituted for y the equation is satisfied. An equation containing only first derivatives or differ- entials is called an equation of the first order. In general, the order of a differential equation is the order of the highest derivative occurring in it. 2. Separation of the Variables. — If a differential equa- tion has the form fi(x)dx+f2(y)dy = 0, (2a) 1 2 DIFFERENTIAL EQUATIONS Chap. I one term containing only x and dx, the other only y and dy, the variables are said to be separated. The solution is ob- tained by integration in the form Jfi (x) dx + Jh iv) dy = c, (2b) where c is a constant of integration. Since the integration formulas contain a single variable, if the variables are not separated, we cannot solve the equation in this way. Thus, if xdy -\- {1 — y) dx = 0, since x dy cannot be integrated, we cannot obtain a solution by direct integration. By division we can however reduce this equation to the form ^ + •^-^=0 (2c) I -y x in which the variables are separated. The solution is then In a; — In (1 — y) = c. When the variables can thus be separated the differential equation is called separable. An equation of the form Mdx + Ndy = is separable when each of the coefficients M, iV is a function of only one variable or the product of factors each contain- ing a single variable. 3. Different Forms of Solution. — The solution In a; - In (1 - 2/) = c (3a) In: can be written whence 1-2/ 1-2/ ' = e' = k. Chap. I FIRST ORDER EQUATIONS 3 Since c is an arbitrary constant, k is also arbitrary. The solution could then be written x = c(l-y) (3b) where c is an arbitrary constant. It could also be written 1 — y = ex (3c) or y - 1 = ex. (3d) Any one of the equations (3a), (3b), (3c), (3d) is the solution of (2c), but of course the constant has a different meaning in each case and so two of these could not be used simultane- ously. Example. Solve the equation (1 + x^) dy — xy dx = 0. Separating the variables, this becomes dy xdx _ ^ whence In 2/ — § In (1 + x^) = const. Since any constant is the logarithm of another constant, this can be written ' In 2/ — I In (1 + x^) = In c, whence y = cVl + x^. This answer could equally well be written in any one of the forms 2/2 = c^ (1 + a;2), y' = c{l+ X'), q/2 = 1 + x^. EXERCISES Solve the following equations: 1. tan X sia^ y dx + cos'' x cot y dy — 0. 2. {xy'' +x)dx + {y — x'y) dy = 0. DIFFERENTIAL EQUATIONS Chap. I 3. {xy^ + x) dx + {x^y — y)dy = 0. • , dy 4. tan x~- — y = a. dx en @+0= '■■£+'-'•■ 4. Derivative Relations. — In many cases one or more of the quantities occurring in a problem is a derivative. An equation satisfied by these quantities is then an equation containing a derivative, i.e., a differential equation. Thus it may be known that the slope of a curve is a given function of x and y. Since the slope is -^ > the curve can be obtained by solving the differential equation Again, it may be known that the velocity oi a moving particle is a given function of the distance s and time t. The differential equation is then More generally, if the rate of change of a quantity x is known to be a function/ {x, t), then 5. Determination of Constants. — Since the constant of integration may have any value whatever, there are an infinite number of solutions of a given differential equation. A pair of corresponding values of the variables is however usually known. By substituting these in the solution the constant can be determined and so a definite solution be ob- tained. In many cases the derivative is known merely to be pro- Chap. I FIRST ORDER EQUATIONS portional to a cer-tain function / {x, y). The differential equation is then dy dx = kfix,y), where k is constant. If two pairs of corresponding values Xi, 2/1 and Xi, t/2 are known, by substituting in the solution both k and c can be determined. The statement of a problem thus consists of two parts. One part contains conditions true at all places or times. From this the differential equa- tion is determined. The second part contains conditions true at a single place or time. These are used to determine the con- stants. Example 1. Find the curve passing through (2, 3) such that the part of the tangent between the coordinate axes is bisected Fig. 5. at the point of tangency. Every tangent is bisected at the point of tangency. Let P (x, y) be the middle point of the tangent AB. Then by similar triangles OA = 2y, OB = 2x. The slope of the curve at P (x, y) is dy dx OA OB This can be written dx &y ^ Q x y ' the solution of which is xy = c. 6 DIFFERENTIAL EQUATIONS Chap. I Since the curve passes through (2, 3), we must then have 2 (3) = c. Hence the equation of the curve is xy= 6. Example 2. Radium decomposes at a rate proportional to the amount present. If half the original quantity disap- pears in 1600 years, what percentage disappears in 100 years? Let R be the amount of radium present at time t. The rate of decomposition is measured by — -5- • Since this is proportional to R, ^-kR dt ~ "^^ where k is constant. Hence and -^ = kat a In 22 = fci + c. Let ir!o be the amount at the start. Substituting t = 0, R = Ro, we have In Ro = c. Substituting this value of c and transposing, we have In-^ =kt. When t = 1600, R = ^ Ro. Hence In I = 1600 k, whence h= In 2 1600 ■ Chap. I FIRST ORDER EQUATIONS 7 When t — 100 we therefore have 1^0 = - 1™ : ^«' = - -^^^^ which gives ^ = .958. This shows that 95.8% remains at the end of 100 years and so 4.2 % disappears. 6. Differential Relations. — It is usually easier to find relations between the first differentials of the variables than between the variables themselves. This is due to certain simplifying assumptions that may be made without affecting the results. Thus, so far as first differentials are concerned, a smaU part of a curve near a point may be considered straight and a part of a surface plane; during a short time dt a particle may be considered as moving with constant velocity and any physical process as occurring at a con- stant rate. The reason these assumptions give a correct result is because the ratio of differentials is by definition the limit of the ratio of increments, and as the increments approach zero these simple conditions become more and more approximately satisfied. Methods of setting up differential relations in this ap- proximate way are often called differential methods. As here stated these methods apply only to first differentials, or first derivatives. A correct equation containing second derivatives would not in general be obtained by considering a small part of a curve as straight and in a small interval a physical process as occurring at a constant rate. Example 1. Find the shape of a mirror such that all fight coming from one fixed point is reflected to another fixed point. Let the fight from F (Fig. 6) be reflected to F'. The mirror must have the form of a surface of revolution. Other- 8 DIFFERENTIAL EQUATIONS Chap. I wise light passing out from F in a plane through FF' would not be reflected in the same plane and so could not go to F'. Let PQ be an infinitesimal arc. With F as center construct the arc QS and with F' as center the arc QR. Consider PQS and PQR as right triangles. They have the common h3^o- tenuse PQ. Also, since the angles of incidence and reflection are equal, Z QPS = z QPR. Fig. 6. Hence the right triangles are equal and so PS = PR. (6a) Let r = FP, r' = F'P. In passing from P to Q the in- crease in r is dr = PS and that in r' is dr' = - PR. Hence, from (6a), dr — — dr' and so r + r' = const. (6b) The section of the mirror by any plane through FF' is there- fore an elhpse with F, F' as foci. Example 2. The sum of $100 is put at interest at 5% per annum under the condition that the interest shall be com- pounded at each instant. How many years will be re- quired for the amount to reach $200? Let A be the amount at the end of t years. In the short time dt the increase will equal the interest dA = .05 A dt. Chap. I FIRST ORDER EQUATIONS Integrating between the limits A = 100 and A = 200, we get '^^^=.05^. fioo A /•2( J 100 whence , 1 1 200 ,_„ < = -Qgln j^ = 13.9 years. 7. Flow of Water from an Orifice. — If there were no loss of energy, the velocity with which water would issue from an orifice at depth h below the surface would be that acquired by a body in falhng the distance h, namely, V2gh. Because of friction and the converging form of the stream, the average velocity with which the water actually issues is V = c V2 gh, where, for ordinary small orifices with sharp edges, c = 0.6 ^ --.^^ approximately. \^^^;-«ai ^----.-.-i-r:-_--_-.-- ^ Example. Find the time v^ r r ~~f required for a hemispherical ^^"====4==^==^ bowl 2 ft. in diameter to empty \^^ jh ^/'^ through an inch hole at the bottom. Let h be the depth of water at time t and FiQ. 7. r = Vi - (1 - hy the radius of the circle forming its surface. The water which issues in time dt generates a cylinder of altitude v dt with an inch circle as base. Its volume is '^QT" dt. 10 DIFFERENTIAL EQUATIONS Chap. I This causes the loss from the surface of a slice of water with radius r, altitude — dh, and volume -Tr^dh = Tr{¥ -2 h) dh. Hence 7r(/i2 -2h)dh = ^('^Yy \.,t = .^AZ^dt (24) Separating the variables, we have t = 120 r {U - 2 U) dh = 112 sec. 8. Equation of Continuity. — In a physical process there may occur an element which is neither created nor destroyed. The amount of this element in a given region only changes when some comes in or goes out through the boundary. In such a case the obvious equation increase = income — output is sometimes called an equation of continuity. Stated in differential form this may give a differential equation by which the variation of the particular element can be de- termined. The concentration c of a particular substance is the amount of that substance in unit volume. If the concentration is uniform the amount in the volume v is then cv. Example. In a tank are 100 gallons of brine containing 50 lbs. of dissolved salt. Water runs into the tank at the rate of 3 gals, per minute and the mixture runs out at the same rate, the concentration being kept uniform by stir- ring. How much salt is in the tank at the end of one hour? Let X be the amount of salt in the tank at the end of t minutes. The concentration is then c = yTTpj pounds per gallon. Chap. I FIRST ORDER EQUATIONS 11 In the time dt, 3 dt gals, of water come in and 3 dt gals, of brine containing Z cdt lbs. of salt go out. Hence the change in the amount of salt in the tank is dx = — 3 c di = — -^ dt. The amount of salt at the end of one hour is then deter- mined by whence and /""dx _ __3_ /•«» ^x~ 100 jo ' ^5-0= -1-^ = 8.27 lbs. 9. Flow of Heat. — If the temperatures at the bounding surfaces of a body are kept constant, the body will ulti- mately approach a steady state in which the temperatures at different points may be different but the temperature at a given point no longer changes with the time. In many cases the temperature T is a function of a single coordinate X. By Newton's law, the rate at which heat flows across an area A perpendicular to x is then -hA^=Q, (9a) where fc is a constant called the conductivity of the material. If we have a series of surfaces A such that the heat flow- ing across one flows across all the others, the equation of continuity has the form Q .= const. (9b) If then A is expressed in terms of x, the solution of (9a) gives r as a ftinction of x. By substituting the values of X and T at two boundaries, the constant of integration and Q can then be determined. 12 DIFFERENTIAL EQUATIONS Chap. I Example. A hollow spherical shell of inner radius 6 cm. and outer radius 10 cm. is made of iron (k = 0.14). If the inner temperature is 200° C. and the outer 20° C, find the temperature at distance r from the center and the amount of heat per second that flows outward through the shell. By symmetry the flow of heat is seen to be radial. At distance r from the center the area across which heat is flowing is the spherical surface A = 4 Trr^ Fig. 9 Since there is no accumulation of heat between the surfaces, the same amount flows across each spherical surface and so equation (9) is dT — Airkr^-y- = Q = const. dr Separating the variables and integrating, we get 4:^kT = ^+c. r Substituting T = 20, r = 10 and T = 200, r- = 6, we find C = - 1000 Trk, Q = 10,800 wk, r = 2700 _ 250. The rate of flow through the shell is Q = 10,800 Tvk = 4750 cal./sec. 10. Second Order Processes. — In a problem con- taining two independent variables x and y it is sometimes stated that a quantity z is proportional to x and also pro- portional to y. What is meant is that, y being constant Chap. I FIRST ORDER EQUATIONS 13 z is proportional to x, and x being constant z is proportional to y. Both statements are expressed by the equation z = kxy. Thus, tne rate at which a substance x dissolves is pro- portional to the amount of x present. It is also propor- tional to the difference between c, the concentration of x in the solvent, and s its concentration in a saturated solution. These statements are both expressed by -jT = kx (s — c). (10a) The fact that the total amount of x present (soHd and in solution) is constant gives an equation of continuity from which we can express c in terms of x and so express 37 as a quadratic function of x. A process in which the rate of change of a; is a quadratic function of x is called a second order process. Example. Sulphur is to be removed from an inert material by extraction with benzol. By using a large amount of benzol it is found that half the sulphur can be extracted in 42 min. If the material contains 6 gms. of sulphur and 100 gms. of benzol are used, which if saturated would dissolve 11 gms. of sulphur, how much of the sulphur wiU be removed in 6 hrs.? Let X be the amount of sulphur undissolved at time t. The concentration of sulphur in a saturated solution is s = :7^gms. sulphur per gm. benzol. The differential equation for x is then ^ = fe (0.11 - c). (10b) 14 DIFFERENTIAL EQUATIONS Chap. I If we use a very large amount of benzol, c will be very small and so this may be replaced by ^=fcx(0.11). at Since x then varies from 6 to 3 in 42 min., T — = 0.11 & Hdt, Jb X Jo whence k = - 0.15. If now we use 100 gms. benzol, when there are x gms. sulphur undissolved, there will be 6 — a; in solution and so 6 - X 100 Hence (10b) can be written dx It and 0.15 ^('o.ll - \qq^) = - .0015 X (.5 + x), '360 dt XX dx r- , , ,s = - .0015 / a; (x + 5) Jo ^^6FT5)=-2-'' which gives x = .19 gms. as the amount of sulphur undis- solved at the end of 6 hrs. PROBLEMS 1. Find the equation of the curve passing through the origin such that the part of every tangent between the a>-axis and the point of tangency is bisected by the y-axis. 2. Find the curve passing through the point (2, 0) such that the part of the tangent between the y-axis and point of tangency is of length 2. J-' 3. If in the culture of yeast the amount of active ferment doubles Chap. I FIRST ORDER EQUATIONS 15 in one hour, how much may be anticipated at the end of 2J hours at the same rate of growth. 4. If the activity of a radioactive deposit is proportional to its rate of diminution and is found to decrease to j its initial value in 4 days, find the value of the activity as a function of the time. 5. The retarding effect of fluid friction on a rotating disk is pro- portional to its angular velocity. If the disk starts with a velocity of 100 revolutions per minute and revolves 60 times during the first minute, find its velocity as a function of the time. y" 6. According to Newton's law the rate at which a substance cools in air is proportional to the difference of its temperature and that of the air. If the temperature of the air is 20° C. and the substance cools from 100° to 60° in 20 min., when will its temperature become 30°? 7. A substance is xmdergoing transformation into another at a rate " proportional to the amount of the substance remaining un transformed. If that amount is 31.4 at the end of 1 hr. and 9.7 at the end of 3 hrs., , find the amount at the start and find how many hours wiU elapse before only 1% wiU remain. 8. When a Hquid rotates about a vertical axis, show that its surface forms a paraboloid of revolution. Observe that the weight of a par- ticle of water at the surface and its centrifugal force must have as resultant a force perpendicular to the surface. 9. Through each point of a curve lines are drawn parallel to the axes to form a rectangle two of whose sides lie on the axes. Find the curve which cuts every rectangle of this kind into two areas one of which is twice the other. 10. Find the surface of a mirror such that all Ught from a fixed point is reflected parallel to a fixed line. Take the fixed point as or- igin and let the light be reflected parallel to the x-axis. Use the polar coordinate r and the x-coordinate as variables. 11. In a certain type of reflecting telescope light converging toward a fixed point is reflected by a mirror to another point. Find the shape of the mirror. 12. If a man can earn 5 dollars per day over expenses and keep his earnings continuously invested at 6% compound interest, how long will it take to save $25,000? 13. If a man can earn s dollars per year over expenses and keep his savings continuously invested at 6% compound interest, how long will be needed to obtain an income of s dollars per year from investments? 14. The amount an elastic string of natural length I stretches under a force F is klF, h being constant. Find the amount it stretches when suspended from one end and allowed to stretch under its own weight w. 16 DIFFERENTIAL EQUATIONS Chap. I 15. Find the amount the string of the preceding problem stretches if it is hung up with a weight P attached to the lower end. 16. Consider a vertical column of air and assume that the pressure at any level is due to the weight of air above. Find the pressure as a function of the height if the pressure at sea level is 14.7 lbs. persq. in., and at an elevation of 1600 ft. is 13.8 lbs. per sq. in. Assume Boyle's law that the density of the gas is proportional to the pressure. 17. If air in moving from one level to another expands without receiving or giving out heat, that is, adiabatically, p = fcp", where p is the pressure, p the density, and k, n constants. Assuming adia- batical expansion, if the density at sea level is .08. lbs. per cu. ft. and the pressure 2100 lbs. per sq. ft., find the height of the atmosphere. 18. If the coefficient of friction between a belt and pulley is jtt and the angle of lapping a, show that the tensions 7'i, Ti in the two sides of the belt when it is slipping satisfy the equation 19. If the velocity is high centrifugal force reduces the pressure of the belt upon the pulley. Assuming the weight of the belt to be w lbs. per unit length and its velocity v, find the equation connecting the tensions in the two sides of the belts. 20. The end of a vertical shaft of radius o is supported by a flat- step bearing. If the horizontal surface of the bearing carries a uniform load of p lbs per sq. in. (new bearing) and the coefiBcient of friction is fi, find the work done against friction in one revolution. An old bearing is worn a little more at the edge than at the center. Ultimately the pressure varies in such a way that the wear is the same at all points. Assuming that the wear at any point is proportional to the work of friction per unit area at that point, find the law of vari- ation of pressure and show that with a given total load the work per revolution is only f that in a new bearing. 21. Assuming that the density of sea water Under a pressure of p lbs. per sq. in. is 1 + 0.000003 p times its density at the surface, show that the surface of an ocean 5 miles deep is about 450 ft. lower than it would be if water were incom- pressible. 22. A cylindrical tank with vertical axis is 6 ft. deep and 4 ft. in diameter. If the tank is full of water, find the time required to empty through a 2-inch hole at the bottom. Chap. I FIRST ORDER EQUATIONS 17 23. Find the time of emptying if the axis of the tank in the preceding problem is horizontal. 24. Two vertical tanks each 4 ft. deep and 4 ft. in diameter are connected by a short 2-inch pipe at the bottom. If one of the tanks is fuU and the other empty, find the time required to reach the same level in both. Assume that the velocity through the pipe is the same as that through an orifice under the same effective pressure. 25. Into a tank of square cross-section, 4 ft. deep and 6 ft. in diam- eter water flows at the rate of 10 cu. ft. per minute. Find the time required to fill the tank if at the same time the water leaks out through an inch hole at the bottom. 26. If half the water runs out of a conical funnel in 2 min., find the time required to empty. 27. A vertical tank has a sUght leak at the bottom. Assuming that the water escapes at a rate proportional to the pressure and that ^ of it escapes the first day, find the time required to half empty. -r- 28. In a tank are 100 gals, of brine containing 50 lbs. of dissolved salt. Water runs into the tank at the rate of 3 gals, per min., and the mixture runs out at the rate of 2 gals, per min., the concentration being kept uniform by stirring. How much salt is in the tank at the end of one hour? -. 29. Suppose the bottom of the tank in the preceding problem is covered with a mixture of salt and insoluble material. Assume that the salt dissolves at a rate proportional to the difference between the concentration of the solution and that of a saturated solution (3 lbs. salt per gal.) and that if the water were fresh 1 lb. salt would dissolve per minute. How much salt wiH be in the solution at the end of one hour? 30. Oxygen flowsithrough one tube into a hter flask filled with air and the mixture of oxygen and air escapes through another. If the action is so slow that the mixture in the flask may be considered uni- form, what percentage of oxygen wiU the flask contain after 5 liters have passed through? Assume that air contains 21% oxygen. 31. The air in a recently used class-room 30' X 30' X 12' tested 0.12% carbon dioxide. How many cu. ft. air containing 0.04% CO2 must be admitted per minute that 10 minutes later it may test 0.06% 062. 32. If the average person breathes 18 times per minute exhaling each time 100 cu. in. containing 4% CO2, find the per cent CO2 in the air of a class-room j hour after a class of 50 enters, assuming the air fresh at the start and that the ventilators admit 1000 cu. ft. fresh air per minute. Let the volume of the room be 10,000 cu. ft. and assume that fresh air contains 0.04% COj. 18 DIFFERENTIAL EQUATIONS Chap. I 33. A factory 200' X 45' X 12' receives through the ventilators 10,000 cu. ft. fresh air per minute containing 0.04% COz. A half hour after the help enters at 7 a.m. the CO2 content has risen to 0.12%. What value is to be anticipated at noon? 34. A brick wall (fc = 0.0015) is 30 cm. thick. If the inner surface is at 20° C. and the outer at 0° C, find the temperature in the wall as a function of the distance from the outer surface. Also find the heat loss per day through a square meter. 36. A steam pipe 20 cm. in diameter is protected with a covering 10 cm. thick of magnesia (k = 0.00017). If the outer surface is at 30° C. and the surface of the pipe 160° C, determine the temperature in the covering as a function of the distance from the center of the pipe. Also determine the heat loss per day through a meter length of the pipe. 36. A wire whose resistance per cm. length is 0.1 ohm is imbedded along the axis of a cylindrical cement tube of radii 0.5 cm. and 1.0 cm. An electric current of 5 amp. is found to keep a temperature difference of 125° C. between the inner and outer surfaces. What is the conductivity of the cement? 37. The amount of light absorbed in passing through a thin sheet of water is proportional to the amount falling on the surface and also proportional to the thickness of the sheet. If one-half the light were absorbed in penetrating 10 ft., how much would reach the depth of 100 ft.? 38. A porous material dries in a confined space at a rate proportional to its moisture content and also to the difference between the moisture content of air and that of saturated air. A quantity of material con- taining 10 lbs. of moisture was placed in a closed storeroom of volume 2000 cu. ft. The air at the beginning had a humidity of 25%. Sat- urated air at the given temperature contains approximately 0.015 lbs. moisture per cu. ft. If the material lost half its moisture the first day, estimate its condition at the end of the second day. 39. How long would be needed for the substance of the preceding problem to lose 90% of its moisture if the humidity of air is kept at 25% by ventilation. 40. A mass of insoluble material contains 30 lbs. of salt in its pores. The mass is agitated with 30 gals, of water for 1 hour when one-half the salt is found to be dissolved. How much would have dissolved in the same time if we had used double the amount of water? Assume the rate of solution proportional to the amount of undissolved salt and also proportional to the difference between the concentration of the solu- tion and that of a saturated solution (3 lbs. salt per gal.). 41. A mass of inert material containing 5 lbs. of salt in its pores is agitated with 10 gals, of water. In 5 minutes 2 lbs. of salt have dissolved. When wiU the salt be 99% dissolved? CHAPTER II OTHER FIRST ORDER EQUATIONS 11. Exact Differential Equations. — The equation Mdx + Ndy = (11a) is called exact if their exists a function u with total differential du = Mdx + N dy. (lib) In this case (11a) can be written du = and so its solution is u = c. It is shown in calculus that M dx + N dy is an exact differential when and only when BM dN .^, , ~E— = ^— • (lie) dx dy To determine whether an equation is exact we therefore calculate these partial derivatives and observe whether they are equal. To solve the equation it is necessary to find the function u whose differential is M dx + N dy. The terms can often be arranged in groups each of which is an exact differential. The value of u is then obtained by integrating these groups separately. If this cannot be done, the solution can be determined from the fact that ^ = M. dx 19 20 DIFFERENTIAL EQUATIONS Chap. II By integrating with y constant, we get u =jMdx+f{y), the constant of integration being possibly a function of y. This function of y can be found by equating the differential oi u to N dx + N dy. Since df (y) gives terms containing y only, / (y) can usually be found by integrating the terms in N dy that do not contain x. In exceptional cases this may not give the correct result. The answer should there- fore be tested by differentiation. Example 1. (x + y) dx + (2 y + x) dy = 0. This equa- tion can be written xdx + 2y dy + (ydx + xdy) =0. It is therefore exact and its solution is x^ + 2y^ + 2xy = c. Example 2. e" dx + (xe^ — 2y) dy = 0. In this case dy dy^ ' dN d , ^~=-(^e^-2y)=e^. These derivatives being equal, the equation is exact. Hence M = I e^ dx = xe^ + f (y), du = e" dx + \xe" + f{y)\ dy, where f{y) is the derivative of / {y). Comparing with the original equation, we see that /'(?/)= -2y and so f(y)=- 1. Chap. II OTHER FIRST ORDER EQUATIONS 21 The solution is xe^ — y^ = c. The value of / {y) could have been obtained by integrating — 2ydy which is the part of {xey -2y)dy not containing x. 12. Integrating Factors. — If an equation of the form M dx + N dy = is not exact it can always be made exact by multiplying by a proper factor. Such a multipher is called an integrating Jador. For example, the equation 2/ (1 + xy) dx — xdy = is not exact. It can however be written y dx — xdy + xy^ dx = 0. Dividing by y^, y dx — X dy r + X dx == 0. Both terms of this equation are exact differentials. The solution is y + 2'' ='• An integrating factor in this case is -j • While an equation of the form M dx + N dy = always has integrating factors, there is no general method of finding them. 13. Linear Equations. — A differential equation of the form ^ + P2/ = Q, (13a) 22 DIFFERENTIAL EQUATIONS Chap. II where P and Q are functions of x or constants, is called linear. A linear equation is thus one of the first degree in one of the variables {y in this case) and its derivative. Any functions of the other variable may occur. When the equation is written in the form (13a), fPdx e is an integrating factor; for, when multipKed by this factor, the equation becomes e % + ye P=e Q. The left side is the derivative of fPdx ye Hence fPdx „ fpdx ye = je Q dx + c (13b) is the solution. Example 1. -^ + — y = 3?- In this case fpdx= f-dx = 2\nx == In x\ Hence fPdx Inx2 e = e = x^. The integrating factor is therefore x'. Multiplying by x^ and changing to differentials, the equation becomes x^ dy + 2 xy dx = x? dx. Chap. II OTHER FIRST ORDER EQUATIONS 23 The solution is 3?y = ga;« + c. Example 2. (1 + 'jp) dx — {xy -\- y -\- rf) dy = 0. This is an equation of the first degree in x and dx. Dividing by (1 + y') dy, it becomes dx y _ d^'TTJ'"^'^' P is here a function of y and fPdy J e = , multiplying by this factor, the equation becomes dx xy dy _ y dy whence and x = l + y^ + c VlTW- 14. Equations Reducible to Linear Form. — An equation of the form ^ + Py = Qy" (14) where P and Q are functions of x can be made Unear by a change of variable. Dividing by y^it becomes y^f^ + Py--+' = Q. 24 DIFFERENTIAL EQUATIONS Chap. II If we take 2/'~" = u as new variable the equation becomes 1 du 1 — ndx which is Knear. Example. -^ + -y = ^- ax X 3? + Pu = Q, Division by y^ gives Let Then -sdy ,2 _2 ^ 1^ ^ dx^x^ a?' u = y-^. whence £ = -^^l"' ^ dx 2dx Substituting tnese values we get and so Idu 2 ^1^ 2dx X cc' du 4 dx X : This is a linear equation with solution 1 3x^ w = qV2 + ^' or, since u = y~', Chap. II OTHER FIRST ORDER EQUATIONS 25 15. Homogeneous Equations. — A function / (x, y) is said to be a homogeneous equation of the nth degree if J{tx,ty) = t»fix,y). Thus, VxM-^is a homogeneous function of the first degree; for Vx^ f + 2/2 e = t Vx^ + y^. It is easily seen that a polynomial all of whose terms are of the nth degree is a homogeneous function of the nth degree. The differential equation M dx + Ndy = is called homogeneous if M and N are homogeneous func- tions of the same degree. To solve a homogeneous equation, substitute y = vx, or x = vy. The new equation will be separable. Example 1. x-^ — y = Va;^ + y^. This is a homogeneous equation of the first degree. Sub- stituting y = vx, it becomes X (v + X -J-) — vx = Vx^ + 2)2 x^, whence .'i.VTT7. This is a separable equation with solution X = c(v + Vl + 1)2), 26 DIFFERENTIAL EQUATIONS Chap. II Replacing zj by - j transposing, squaring, etc., x^ — 2cy = &. E.arnvle2. y (^J + 2 J/^- y = 0. Solving for -^ > we get dy _ — X ± y/x^ + if dx y whence y dy + xdx = db Va;^ + y^ dx. This is a homogeneous equation of the first degree. It is however much easier to divide by Vx^ + y^ and integrate at once. The result is xdx + y dy , — ' ^ ^ = ± dx. Vx^ + 2/2 Integration gives Vx' + y^ = c ± X and so J/2 = g2 _j_ 2 ex. Since c may be either positive or negative, the answer is equivalent to ^2 _ g2 _j_ 2 ex. 16. Change of Variable. — We have solved the homo- geneous equation by taking X as new variable. It may be possible to reduce any equa- tion to a simpler form by taking some function m of a; and y as new variable or by taking two functions u and v as new variables. i, Chap. II OTHER FIRST ORDER EQUATIONS 27 If the differential equation only is known some expression appearing in the equation may be a good variable. Thus it often happens that y appears only in the combinations y^ and y —- • By taking , dy Idu ^="' ydrx = 2Tx' a simpler equation is obtained. If the equation is obtained in the solution of a problem, any quantity which plays a prominent role in the statement of the problem may be a good variable. Thus, in solving the reflector problem (Art. 6) we used as variables the dis- tances from the two points which were directly suggested by the problem itself. dv Example, (x — yY -j- = a?. . Let X — y = u. Then ^ _dy _du dx dx and the differential equation becomes n O •> C^ ti u^ — a' = u^ T-' dx The variables are separable. The solution is , a, u — a , whence a, X — y — a , y = jrln ^-5— + c. " 2 X — y + a 28 DIFFERENTIAL EQUATIONS Chap. H 17. Simultaneous Equations. — We often have two dif- ferential equations containing two dependent variables x and y and their de- rivatives with respect to the same independent variable t. It may be possible to combine the equations algebraically so as to get an equation containing only one dependent variable, which may be a; or y or any function of x and y. We solve for this and substitute in one of the original equations to complete the solution. Example 1. -^ = ki{x — y),-n = hy. The second equation contains only one dependent variable y. Its solution is y = cie*^. Substituting this value in the first equation, -rr- — kix = — fciCie*^' at ^ This is a linear equation with solution X = c^e '^ — 5 ^ e '^ • Ki — fci Example 2. -n = ^ — V) 737 = 2 y. Multiplying the first equation by a constant a, the second by b and adding, we get ^l (ay + bx) = ax+ (2 b - a) y. The right side of this equation will be a multiple of the expression in parentheses if a _ 2b — a b a Chap. II OTHER FIRST ORDER EQUATIONS 29 A solution of this is a = b = 1, and so Solving for x + y, x + y = cie*. Substituting X = cie' - y (17) in the first equation we get This is a Unear equation with solution y = de-^ +gCie'. Substituting in (17), we get X = -xCie? — der^. EXERCISES Solve the foUowing differential equatioBs: 1. (3x^ +2xy - y^) dx + (a? -2xy -3 y^) dy = 0. -2. x-r- +y = 3?. ax Jbt-^- {x^ + y^) dx+2xydy = 0. ^i. {x^ + 2/2) dx —2xydy =^ 6. ydx + (x+ y) dy = 0. 6. ydx — {x + y) dy = 0. >^. xdy +y dx =3/" dx. 8. 'i-ay=^^. 30 DIFFERENTIAL EQUATIONS Chap. U 9. x^g-2.2/ = 3. 10. x^^ - 2x2/ = 3 2/. 11. (2 X2/2 - y)dx +xdy =0. 12. tan X ~ — 2/ = a. dx 13. (x2 - 1)J dy + (x^ + 3 X2/ v'i^'^) dx = 0. 14. ye" dx = (j/' + 2 xe*) dy- 15. (xy ei + 2/^) dx — 3?ei dy = 0. 16. 1 + 2/ = 0=2/3. 17. x^-3 2/ +x<2/^ = 0. dx 18. xdx -'r y dy = xdy — y dx. 19. (x2/^ — x) dx + (2/ + xy) dy = 0. 20. xdy — y dx = Vx* + 2/^ dx. 21. (sin X + y) dy + (y cos x — x'') dx = 0. 22. xdy — ydx = X Vx^ + 2/^ dx. 23. (1 + x^) dy + {xy - x^) dx = 0. 24. y dx = iy^ — x) dy. 26. y ^ + 2/* cot X = cos x. dx 26. (x + 2/ - 1) dx + (2 X + 2 2/ - 3) di/ = 0. 27. 3 2/^^-2/3 =x. 29 30. 0/1^+1] =&:, 31. (2 X + 3 2/ - 1) dx + (4 X + 6 2/ - 5) d2/ = 0. 32. (3y'+3xy+ x") dx = (x=i + 2 xy) dy. 33. dx , , dy ^+x = e'. J? = x. 34. J+x = l. ^ + 2/ = l. 36. about its axis. Ultimately the mass of gas will rotate hke a rigid body. Assuming Boyle's law and taking account of centrifugal force, find the law connecting the pressure of the gas and the distance from the axis. Would a similar law be obtained, if instead of a cylinder, the container had any other shape? 6. In a chemical reaction a substance c decomposes into two sub- stances X and y, the rate at which these products are formed being pro- portional to the amoimt of c present. If at the beginning c = 1, X = Q, y = Q, and at the end of 1 hr. c = |, a; = |, 2/ = f, find x and y as functions of the time. 7. In a certain chemical reaction, 1 mol. of y is produced for each mol. of X consumed, the rate being proportional to the amount of x present, and at the same time yhj a. reverse reaction is converted into a; at a rate proportional to the amount of y present. Chemical analysis showed t =0, 3, 00 X = 10, 6, 5.5 2/ = 0, 4, 4.5. Find X and y as functions of t. 32 DIFFERENTIAL EQUATIONS Chap. II 8. Radioactive substances are transformed at a rate proportional to the amount of substance present. Through the decomposition of a mol. oi Ra B sk mol of Ra C is produced, the rate being such that J the Ra B disappears in 27 minutes. Similarly the Ra C decomposes at a rate such that J of it is lost in 19.5 min. If initially 1 mol. of Ra B is present, find the amount of Ra B and Ra C at the end of 1 hr. 9. When light passes from a medium of refractive index ^ to one of index aj', fi' _ sin y. sin B' 6 and B' being the angles which the incident and refracted ray make with the normal to the surface of separation. According to Einstein's theory, the gravitational field of the sun deflects a ray of light as if it had a refractive index. - 1 -u" where a is constant and r the distance from the center of the svm. Find the path described by the ray. 10. Suppose bacteria grow at a rate proportional to the number present but that they produce toxines which destroy them at a rate proportional to the number of bacteria and to the amount of toxin. Suppose further that the rate of production of toxin is proportional to the number of bacteria. Show that the number increases to a maxi- mum and then decreases to zero and at time i is given by 4M N = (ew -I- e-t'y where M is the maximum number and t is measured from the time when the number is a maximum. CHAPTER III SPECIAL TYPES OF SECOND ORDER EQUATIONS 18. Equations Immediately Integrable. — An equation of the form g = /(x) (18) can be solved directly by two integrations. The first integration gives A second integration gives the general solution in the form -Id' f (x) dx\dx-\- C]X + Co. p=|. (19b) 19. Equations not Containing y. — An equation not con- taining y can be solved for the second derivative and so reduced to the form %<-%} ^^^■> Take as new variable dx Then (Py _ dp dx^ dx and so (19a) can be written 33 34 DIFFERENTIAL EQUATIONS Chap. II This is a first order equation whose solution has the form V = F {x, ci), where Ci is the constant of integration. Substituting -p for p, we have whence y = I F {x, ci) dx + C2. Example. (1 + a;) -5-^ + t^ = 0. dv Substituting p for -^ we have (1+.)| + P = 0. This is an exact equation with solution (1 + x) p = Ci. dti Replacing p by 3- and separating the variables, we have , Ci dx '^y^T+-x' whence 2/ = Ci In (1 + x) + C2. 20. Equations not Containing x. — An equation not con- taining X can be reduced to the form % = f{y,v). (20a) Substitute ' I = P (20b) Chap. Ill SECOND ORDER EQUATIONS 35 and write the second derivative in the form dx^ dx dydx dy These substitutions bring (20a) to the form This is a first order equation which can be solved for p. dij Replacing p hy -^ the result is a first order equation which can be solved by a separation of the variables. Substituting dy _ d^y _ ^dp dx the equation becomes dx ^' dx" ^dy Dividing by p, yv^ = V^V + P^- ^V 2 1 2/^ = 2/== + ?. The solution of this equation is whence and P I g = J/(y + cO = / ^^ = ^ln^ + c.. (20d) yiy + ci) ci y + ci 36 DIFFERENTIAL EQUATIONS Chap. Ill In solving this problem we divided both sides of the equation by p. This is allowable since p = gives y = c (20e) a solution which contains only one constant of integration whereas we are determining the general solution with two constants of integration. It is to be noted that (20d) is a solution of the original differential equation and that it cannot be obtained by giving special values to the con- stants in (20c). Such a solution is called singular and in some problems might be important. EXERCISES Solve the following differential equations: 1. xg=I+.. ^- d-y- dC ~ s2 ■ 9. (:, + l)g-<, + 2)| + .+2.0. , d's _ a^ Chap III SECOND ORDER EQUATIONS 37 21. Deflection of Beams. — When a beam is bent by- vertical forces as shown in Fig. 21a, the fibers in the upper part are stretched and those in the lower part compressed. There is then a neutral curve AB along which they are neither stretched nor compressed. It is seen from the figure that the amount ds a fiber of natural length s is stretched or compressed is given by the equation ds _ s '^ ~ R' where z is its distance from the neutral curve and i2_is the radius of the circle in which the neutral curve is bent. in the fibers between z and z + dz is T = Ewdz-- = Ewdz.^, s R where E is the stretch modulus of elasticity of the ma- terial and w the width of the beam. Since there is no re- sultant force along the beam ■ — i-— ^^ ds r^ ~ -.,^^ ;=*==Tsr^ ^ ~^ J^\ ^^\ :.( L^ yj^ / / R A-w^/ / Fig 21a By Hooke's law the tension / Ew dz • ^^ -p I wzdz = showing that the neutral curve passes through the center of gravity of each cross section. The moment of the total stress about an axis PQ perpendicular to the beam is M RJ wz^ dz = R (21a) when / is the moment of inertia of the section area about that axis. Let {x, y) be the coordinates of P, x being measured 38 DIFFERENTIAL EQUATIONS Chap. Ill along the beam. Since the curvature is usually very small, the slope -5- is nearly zero and so 1 _ dx^ _ d}y R ~ fTTTwvT* ~ ^' ['+(1)7 approximately. Hence (21a) can be written ^I%=M. (21b) M, called the bending moment, is the moment about P of all forces on one side of P, those acting upward producing positive moments, those downward negative. To find the curve in which a beam bends we determine M as a func- tion of X and solve the differential equation (21b). In the problems considered here the beam is of uniform material and has a constant Fig. 21b. cross section. Hence E and I are constants. Example. Find the deflection of a beam of length 2 I, supported at its ends and loaded with a weight w per unit length. Take the origin at the center of the beam. The total weight supported being 2 Iw, the upward thrust of the sup- port at each end is Iw. Consider forces on the right of the point P. At the end is an upward thrust Iw. Its moment about P is wl (Z — x). The only other force on the section PA is the load w (l — x) between P and A. Since the load is uniformly distributed, this can be considered as acting through its center of gravity at distance i (I — x) from P Its moment is then — w {I - x) • ^ (I - x), Chap. Ill SECOND ORDER EQUATIONS 39 the negative sign being used because the force is downward. The total moment about P is M = wl{l - x) -"^{l - xY = Iw {l^ - x^). The differential equation is therefore Ei% = iwq?-x^). A first integration gives Since the beam is horizontal at the center ax - andsoci = 0. A second integration gives Since we have taken the origin on the curve C2 = 0. The equation of the elastic curve is therefore ^/.=i-(^-f;} The maximum departure of the beam from a straight Hne is at the end, given by placing x = I. This is ordinarily called the deflection of the beam. 22. Equilibrium of a Cable. — Let ABC (Fig. 22) be a perfectly flexible cable fastened at A and C and loaded in any way. Take the x-axis horizontal and the j/-axis through the lowest point B on the curve. Consider the part of the cable between B and the variable point P {x, y) on the 40 DIFFERENTIAL EQUATIONS Chap. Ill curve. This is in equilibrium under the action of three forces: (1) A horizontal tension H at B exerted by the section AB of the cable. (2) A tension T along the tangent at P due to the part PC of the cable. ^ (3) A downward force equal to the load W on the part BP of the cable. The total component of force toward the right must equal that toward the left, and the component of force acting upward must equal that acting downward. Hence r cos <^ = H, r sin = W. Dividing the second equation by the first, we get dy^W dx H Since H is constant, if TF is a known function of x, this can be integrated at once. In some cases W is not known but its derivative can be easily determined. In that case, differentiation gives tan = x: = If ■ (22a) (22b) (22c) dx^ H dx from which we can determine the equation of the curve. The answer contains the constant H. This can be de- termined by substituting the known coordinates of B and the ends of the cable. 23. Motion of a Particle in a Straight Line. — If P is the resultant of all forces acting on a particle of mass m, its acceleration a is given by the equation F = ma. (23a) Chap. Ill SECOND ORDER EQUATIONS 41 If the particle moves along a straight line and s is its distance from a fixed point of the Une, its velocity is v = %, (23b) and its acceleration is dv dh ,_„ , In using these formulas we can measure s in either direc- tion along the Une but must then consider as positive the direction in which s increases. The quantities F, v, and a are positive or negative according as they point in the positive or negative direction thus defined. Example. When a body sinks slowly in a liquid the resistance is approximately proportional to the velocity. If the particle starts from rest, find its motion. Let s, considered positive downward, be the distance the body sinks in t seconds. If m is its mass and g the ac- celeration of gravity, the force of gravity is mg, which is positive since the force is downward. The re- sistance acting upward and being proportional to the ve- locity is — kv. The total force acting on the body is then F = mg — kv. Hence equation (23a) is , dv mg — kv = ma = ""^^ ' Separating the variables and integrating, we get k In (mg — kv) = — —t + c 42 DIFFERENTIAL EQUATIONS Chap. Ill Since v = when < = In (mg) = c. Subtracting from the preceding equation and solving for mg — kv, mg — kv = mg e~m'. Since v = -tt > integration gives mg t — ks = -j- e m' + c. Since s = when t = 0, _ m^g "'IT' Substituting this value and solving for s, 24. Motion of the Center of Gravity. — If M is the total mass of a body or system of bodies and F the resultant of all the forces appUed to it, the equation F = Ma (24) determines the acceleration a of the center of gravity. That is, the center of gravity moves as if the whole mass were concentrated at that point and all the forces applied there. If all parts of the body move in the same direction with the same velocity, this equation determines the acceleration of any point of the body. If the parts of a complex system, such as a chain, all move along the same path with the same speed, and F is the component of force along the path, (24) gives the acceleration along the path. Chap. Ill SECOND ORDER EQUATIONS 43 25. Motion in a Plane. — When a particle of mass m moves in a plane or in space, its acceleration still satisfies the equation F ma but the quantities F and a are vectors, that is, have direc- tion as well as magnitude. To obtain an equation whose terms are numbers we project on any line. If the com- ponent of F along any Une is Fx and the component of a along the same line is a^, then Fx = max. Suppose the particle moves in a plane. Let (x, y) be its rectangular coordinates. The components of acceleration along the axes are ttx = ay = df ' If Fx and Fy are the components of force along the axes, the motion of the particle can then be determined by solving the differential equations m^,=Fx, In problems where the force acts along the hne joining the particle to a fixed point it may be more convenient to use polar coordinates with that point as origin. When the particle is at P its accelera- tion is resolved into a com- ponent Or along OP and a component Oe perpendicular to OP. These are d'y p m^ = Fy (25a) Fig. 25a. Or = %-'$)' --M} (-^' 44 DIFFERENTIAL EQUATIONS Chap. Ill If the force acting on the particle has the component Fr along OP and Fg perpendicular to OP, the motion of the particle can be found by solving the equations mar = Fr, mag = Fe. To prove equations (25b) write X = = r cos d, y = rsinO and calculate a^ = dt^' ay = d'y de (25c) in terms of r and d. These are accelerations along OX and OY. To obtain Or project a^ and Uy on OP and add the results. Similarly, to obtain a^ project on the Hne per- pendicular to OP. Example 1. When an electron of charge e moves with velocity z; in a field of mag- netic intensity H, it is pushed sidewise with a force *-H F = -vH sind) c Fig. 25b. where c is the velocity of Kght and = 90° and F=-vH. c The components of v are dx Jt ' dy dt ' Since F is perpendicular to V and numerically equal to - vH, its components are Fig. 25c. e dy c dt e „dx c dt the algebraic signs being determined by inspection of the figure. The equations of motion are therefore d^x d^y m eH dy c dt eH dx ' dt^ c dt These equations can be immediately integrated giving dx eH , m dy _ eH = —X, ' dt c the constants of integration being determined so that x = 0, y = 0, dx Jt = Vo, 1=0 dt 46 DIFFERENTIAL EQUATIONS Chap. Ill when t = 0. Dividing the second equation by the first dy _ eHx dx cmvo — eHy Clearing fractions and integrating, we get eH (x^ + y^) — 2 cmva y = 0, the constant being zero since y = when x = 0. The electron therefore describes a circle of radius cmvo Example 2. A particle of mass m is attracted toward the origin with the force k If it starts from the point (a, 0) with velocity Vo> - per- pendicular to the X-axis, find the motion. Using polar coordinates, the differential equations are m rdV _ /deVl mk^ Idf^ ^ \dt) j~ r' ' md/^dd\ ^ r dt\ dt) The second equation can be immediately integrated giving .de 'dr'^- At the start r = a and de Chap. Ill SECOND ORDER EQUATIONS 47 Hence Ci = avo and do _(Wo dt~ 7^ Substituting this value in the first equation (Pr _ aW — k^ de 7^ multiplying by ^ dt, dr dh ,^ ah^ — k^ , whence (l)'-w-*'>(i.-^) : integration when r = a. Dividing by dv the constant of integration being determined so that -r: = \dt) ~ r* we obtain ,2 _ jj.2 fdrV ^ ^ \dd) Solving this equation and determining the constant so that r = a when S = 0, we finally obtain the equation of the curve in the form "VaW - fc2 . = a sec - Svo -e 26. Rotation about a Fixed Axis. — Let the body (Fig. 26) rotate about the axis through perpendicular to the plane AOB. The portion of the body can be determined by the angle 8 between the fixed Kne OA and the line OB fixed in the body. The angular velocity of rotation is then .=1 (26a) 48 DIFFERENTIAL EQUATIONS and its angular acceleration " dt df' Chap. Ill (26b) Let F be a force in the plane OAB applied to the body at P- The torque about the axis of rotation due to this force is T = Fl, where I is the perpendicular distance from to the Hne FP- The torque is positive when the force F tends to increase 8. The torque and angular ' P acceleration satisfy the equa- tion B T = la, (26c) where I is the moment of inertia of the body about the axis of rotation and T is the Pjq 26. total torque about that axis of all forces acting on the body. This is analogous to (24), the moment of inertia correspond- ing to mass and torque corresponding to force. 27. Combined Translation and Rotation. — Consider the motion of a rigid body of mass M whose center of gravity moves in a fixed plane and which at the same time rotates about an axis perpendicular to that plane. In the problems solved here the axis is always an axis of symmetry and the forces acting on the body lie in the fixed plane. The axis of rotation will then remain perpendicular to that plane. The motion of the center of gravity is determined by the vector equation F = Ma and the rotation by the equation T = la. (27a) (27b) Chap. Ill SECOND ORDER EQUATIONS 49 In case of a system consisting of two or more rigid bodies moving as just described there is an equation of the form (27a) and one of the form (27b) for each body. The force F is in each case the resultant of all forces acting on the body. It may not be possible to determine at once the force that two bodies of the system exert on each other. Such a force can be represented by a letter. It will be found that there are enough equations to determine these unknown forces as well as the accelerations. Example 1. A cylinder of mass M and radius r rotates about its axis. A cord wrapped around the yig. 27a. cylinder is attached to a mass m which drops vertically. If the cyhnder starts from rest, find the angle turned through in t seconds. Let F be the tension in the cord. The torque about the axis of the cylinder is T = Fr. The moment of inertia of a cylinder is Equation (26c) is then - Mr^ a = Fr or \Mra = F. (1) The forces acting on m are F acting upward and the force of gravity mg acting downward. Its acceleration 50 DIFFERENTIAL EQUATIONS Chap. Ill then satisfies the equation ma =mg — F. (2) If 6 is the angle the cyhnder turns through in t seconds and s the distance m falls s = rd and so d^s cPd or df ^'df a = ra (3) By solving (1), (2), (3) simultaneously, we obtain 2 mg d^e Hence e. = {M + 2m)r df mg t^ {M + 2m)r' Example 2. A sphere of mass M and radius r rolls down a plane which makes an angle ^ with the horizontal. If the coefficient of friction is fi and the inchnation is so great that the sphere shdes, find its angular acceleration about the horizontal axis through its center of gravity. The force of gravity on the sphere can be resolved into two components Mg sin . Chap. Ill SECOND ORDER EQUATIONS 51 The moment of inertia of a sphere is Hence 6 2 tiM gr cos = -r Mr^ a 5 and so 5 iig cos a = 2 r PROBLEMS 1. A beam of length 2 Hs supported at its ends and loaded with a weight W at the middle. Find the deflection. 2. Find the deflection of a cantilever beam of length I, held horizon- tal at one end, and loaded with a weight W at the other end. 3. Find the deflection of a cantilever beam fixed at one end and loaded with a weight w per unit length. 4. Find the deflection of a beam supported at both ends and at the middle point, loaded with a weight w per unit length. 6. Find the deflection of a cantilever beam fixed at one end, sup- ported at the other and loaded with a weight w per vmit length. 6. Find the deflection of a beam fixed at both ends and loaded with a weight w per unit length. 7. Consider a vertical column fixed at the base, of length I, and sup- porting a weight P. Suppose the weight causes the upper end to be displaced the amount a from the vertical. Calculate the bending moment and determine the curve in which the column bends. By substituting the coordinates of the upper end show that the maximum load the column can support is -m EI. 8. Suppose the ends of the column are rounded but are held in the same vertical line. Find the maximum load the column can support. 9. The cable of a suspension bridge supports a bridge of weight w per unit horizontal distance. Neglecting the weight of the cable find the curve in which it hangs. 10. A series of rods of varying length but the same diameter are hung along a cord. The horizontal distances between consecutive rods are equal and their bottoms are in a straight line. Assuming 52 DIFFERENTIAL EQUATIONS Chap. Ill that they are so close together that the load can be considered con- tinuous, find the curve formed by the cord. 11. A cable is supported by its ends and hangs under its own weight. Find the curve in which it hangs. 12. A telegraph wire weighs 173 lbs. per mile. If the poles are 400 ft. apart and the wire sags 10 ft. at the middle, find the tension at the lowest point of the wire. 13. An arch of a masonry bridge supports a horizontal roadbed and is so constructed that the resultant stress at each point of the arch due to the material above is a compression along the tangent. Find the shape of the arch. 14. A particle of mass m moves in a straight line toward a center of force which attracts with the magnitude where r is the distance from the center. If the particle starts from rest at the distance a, find the time required to reach the center. 15. A motor boat weighing 1000 lbs. is moving in a straight line with a velocity of 60 ft./sec. when the motor is shut off. If the resistance of the water is proportional to the velocity of the boat and is equal to 10 lbs. whefl the velocity is 1 ft./sec, how far will the boat move before the velocity is reduced to 25 ft./sec. How long will be required for this reduction in velocity to take place? 16. A particle of mass m moves toward a fixed center of force which repels with a force K'm times its distance from the center. If it starts from the distance a with velocity ka, show that it wiU continually approach but never reach the center. 17. Find the velocity acquired by a body falUng from an indefinitely great distance to the earth, assuming that the force of attraction varies inversely as the square of the distance from the center of the earth. 18. Find the time required for a body to fall to the earth from a distance equal to that of the moon. Take the radius of the earth as 4000 mi. and the distance from the center of the earth to the moon as 240,000 miles. 19. If a hole were bored through the center of the earth, a body falling in it would be attracted toward the center with a force propor- tional to the distance from the center. Find the time required to fall through. 20. A body slides down a rough inclined plane. If the inchnation of the plane is a and the coefficient of friction is fi, determine the motion if the particle starts from rest. Chap. Ill SECOND ORDER EQUATIONS 53 21. Assume the resistance of the air proportional to the square of the velocity. If the velocity of a faUing body is observed to approach the hmiting value 216 ft./sec. and the body starts from rest, find the motion. 22. A particle is projected vertically upward with velocity vo. As- suming that the resistance of the air is k times the square of the velocity, find the velocity with which it returns to the earth. 23. A chain 6 ft. long starts with 1 ft. of its length hanging over the edge of a smooth table. Neglecting friction, find the time required to slide off. 24. A chain hangs over a smooth peg, 8 ft. of its length being on one side and 10 ft. on the other. Find the time required to slide off. 25. Solve the preceding problem if the force of friction is equal to the weight of 1 ft. of the chain. 26. A projectile is fired with a velocity of 2500 ft./sec. in a direction making 45° with the horizontal. Find the highest point reached and the point where it strikes the ground. 27. A projectile is fired with velocity vo at an angle of elevation a. If the resistance of the air is kv, where v is the velocity and k is con- stant, find the equations of motion. 28. A particle is attracted toward the origin with a force propor- tional to the distance. If it starts from the point (o, o) with velocity vo perpendicular to the x-axis, find the path described. 29. Solve, the preceding problem if the particle is repelled with a force proportional to the distance. 30. An electron moves in a magnetic field of intensity H. If it starts with velocity vo in a direction making the angle a with H, find the path described (See Example 1, page 44). 31. Prove equations (25b). 32. Determine the orbit of a planet of mass m assuming that it is attracted toward the sun with the force km where r is the distance from the sun. Let n be its distance and vo its velocity when nearest the sun. 33. Find the orbit of a comet. Let ro be its least distance from the sun and assume that its velocity at an infinite distance is zero. 34. A particle of mass m is attracted toward the origin with the force k'hn u 54 DIFFERENTIAL EQUATIONS Chap. Ill If it starts from the point (a, 0) with velocity h perpendicular to the x-axis, find the path described. 35. A circular disk of radius a submerged in oil rotates about the perpendicular axis through its center. Assume the frictional resistance per unit area at each point of the disk to be kv, where v is the velocity of that point and fc is constant. If the disk is started with angular velocity u and the torque due to friction at the bearings is a constant K, find the motion. 36. A ball of radius r rolls without slipping down a plane. If i^ is the angle of inclination of the plane and the ball starts from rest, find the distance its center moves in t seconds. 37. A billiard ball is started with velocity vo not rotating. If the coefficient of friction between the ball and table is /x, find the motion. 38. A cyhnder of mass M and radius r rolls on the top of a table. A cord wrapped around the cylinder passes horizontally over a fixed pulley and is attached to a weight m which drops vertically. Find the motion of the cylinder. 39. Solve the preceding problem if the cord is attached to the axis of the cylinder. 40. A wedge shaped block of mass M and 45° angle sUdes on a smooth table. A mass m slides on its surface. If both start from rest, find the motion. CHAPTER IV LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 28. Equations of the n-th Order. — The solution of an equation of the n-th order in two variables involves n in- tegrations. The general solution therefore contains n con- stants of integration. Constants of integration are called independent if they occur in the solution in such a way that it is not possible to replace a function of two or more of them by a single constant and so reduce the number of constants. Thus y = cix^ + C2 + CS (28) appears to contain 3 constants. We can however take C2 + cs = c and so obtain y = CiX^ + c. It is not possible to further reduce the niunber. Hence (28) contains two independent constants. In case of a differential equation of the form where / is an algebraic function, it can be shown that there is only one solution containing n independent constants of integration. It is called the general solution. A solution obtained by giving particular values to the constants in the general solution is called a particular solution. Some differential equations have solutions con- taining less than n constants of integration which are not 55 66 DIFFERENTIAL EQUATIONS Chap. IV particular solutions. Such solutions are called singular. They are mainly of mathematical interest and so will not be further considered here. 29. Linear Equations with Constant Coefficients. — A dif- ferential equation of the form d"ii , d"~'« , , dy , , , , ,„„ , is called a linear equation. By this it is meant that the equation is of first degree in one of the variables {y in this case) and its derivatives. If the coefficients ai, oa, . . , On are constants, it is called a linear equation with constant coefficients. For practical appHcations this is one of the most important types. In discussing these equations we shall find it convenient to represent the operation -=- by D. Then ^-Du ^-D'v etc Equation (29a) can be written (D" + ail>"-' + . • • + a„_. I> + a„) 2/ = / (x). (29b) This signifies that if the operation D» + aiZ)"-' + . • . + (z^, I> + a„ (29c) is performed on y, the result is / (x). The operation con- sists in differentiating y, n times, n — 1 times, etc., mul- tiplying the results by 1, ai, a^, etc., and adding. With the differential equation is associated an algebraic equation r" -t- air"- -f • • • + a^-.r + a„ = (29d) having the same coefficients ai, 02, etc. as (29a) but with right-hand member zero. If the roots of this auxiliary equation are r-i, r^, ■ • ■ , r„, the polynomial (29c) can be written (D - ri) (D - ra) . ■ . (Z) - r„) Chap. IV LINEAR EQUATIONS 57 and so (29a) has the form (D - r-i) (D -n) • • ' (D -r„)y=f (x). (29e) If we operate on y with D — ri, we get (D -ri)y = £_- ny. Operating on this with D — r-2 we get (I> - r,) {D -n)y=iD- r,) (g - ny^ The same result is obtained if we operate on y with (D - n) (D - n) = D^ ~ (n + n)D + nr^. Similarly, if we operate in succession with the factors {D — ri), (D — Ti), etc., in any order whatever we get the same result that we should get by operating directly with the product (29c). It should be noted that this is true only when Ti, Ti, etc., are constant. If the r's are variable it is not in general true that (Z) - ri) (D -ri)y={D- r^) {D - n) y. 30. Equation with Right-hand Member Zero. — To solve the equation (Z>» + aiZ)»-' + • • • + a„_, I> + a„) 2/ = 0, (30a) factor the symboUc operator and so reduce it to the form (D - n) (D - ra) • • • (Z) - r„) y = 0. The value y = cie™ is a solution. For (D - 7i) Cie™ = ciTie"" - ncie"' = 0, 58 DIFFERENTIAL EQUATIONS Chap. TV and the equation can be written P-rj) • • . (D-rn) • (D-n) y = {D-r2) • • ■ (D-u) ■ = 0. Similarly y = C2e^'^, y = Cic"', etc., are solutions. Finally y = ae"" + C2e'^ + • • • + c„eV (30b) is a solution; for the result of operating on y is the sum of the results of operating on Cie"^, C2e^"^, etc., each of which is zero. If the roots n, ra, ■ • • ,r„ are all different, (30b) contains n independent constants and so is the general solution of (30a). If, however, two roots n, r2 are equal cie"^ + C2e"^ = (ci + C2)e'''^ contains only one constant Ci + d and (30b) contains less than n independent constants. In this case, however, xe"^ is a solution; for (D — n) xe'"' = nxe'"' + e"^ — rixe'''' = e"^ and so (D - n) (D - Ti) xe'^" ={D - ny xe''^ = (D - n) e"^ = 0. If then 7-2 = n the part of the solution corresponding to these two roots is (ci + Cix)e^'^. More generally, if n = r2 = • • • = /•„, the part of the solution corresponding to these m roots is (ci + caa; + C3a;2 + • • . + c^x^-Oe'"^ (30c) If the coefficients ai, oj, • • 'Oh ar occur in pairs n = a + pV-i, Vi = The terms cie"*, C2e~ Chap. IV LINEAR EQUATIONS 59 Oh are real, imaginary roots are then imaginary but in most problems their sum is real. They can be replaced by two other terms that do not have this imaginary appearance. Using the values of ri and rj we have (D - /-i) {D - rs) = (D - ay + ^•-. By performing the differentiations it can easily be shown that [(D - ay + ^] • e-* sin /3.r = 0, [{D - ay + /S^] • e^ cos fix = 0. Therefore C°" [ci cos fix + d sin 8.v] (30d) is a solution. This function in which a and fi are real can therefore be used as the part of the solution corresponding to two imaginary roots r = a ± /S V^. To solve the differential equation (D" + aiD"-' + . . • + a,_.i) + a,) y = let riffi, ■ • ' , rnbethe roots of the auxiliary equation r» + Oir"-' + • • ••+ a^ r + ch. = 0. // these roots are real and differeivt the solution is y = cic" + cse^ + . . . + f,e'.^ // m of the roots n, ra, • • • r« are equal, the correspoiid- ing part of the solution is (Ci + Csx + cr» + • • • + (WE— ') e"'. 60 DIFFERENTIAL EQUATION Chap. IV The part of the solution corresponding to two imaginary roots r = a ± /3 v — 1 is e"^ [ci cos fix + C2 sin fix]. Examphl. g-|-22/ = 0. This is equivalent to (D2 - D - 2) 2/ = 0. The roots of the aiixiliary equation r2 - r - 2 = are — 1 and 2. Hence the solution is y = cic-^ + C2e*= ExampU2. g + g-5|+3, = 0. The roots of the auxihary equation r^ + r^ -5r + 3 = are 1, 1, — 3. The part of the solution corresponding to the two roots equal to 1 is (ci + C2x)e^. Hence 2/ = (ci + Cix) e + cse"'^. Example 3. (Z)^ + 2 D + 2) ?/ = 0. The roots of the auxiliary equation are - 1 ± -v/^T. Therefore a = - 1, /3 = 1 in (30d) and y = e~^ [ci cos a; + Ca sin x], 31. Equation with Right-hand Member a Function of x. — Let y = u he the general solution of the equation (D" + a.D"-' + • • • + a„-,D + a„) y = Chap. IV LINEAR EQUATIONS 61 and let y = V be any solution of the equation (D- + aiD"-' + • • • + a^,D + a„) y = f (x) (31) then y = u + V is a solution of (31); for when the operation Z)»+ aiD"-' + • • . + a^,D + a„ is performed on u it gives zero and when it is performed on V it gives / (x). Furthermore u + v contains n ar- bitrary constants. Hence it is the general solution of (31). The part u is called the complimentary function, v the particular integral. To solve an equation of the form (31) we first solve the equation with right-hand member zero and then add to the result any solution of (31). A particular integral can often be found by inspection. If not the general form of the integral can be determined by the following rules: 1. If / (x) = ax'' + fex"-' + ■ • • + p, assume y = Ax'' + £x"-' + ■ • ■ +P, but, if occurs m times as a root of the auxihary equation, assume y = x» [Ax"" + 5x"-' + • • ■ + P]. 2. If / (x) = ce^, assume y = A^, but, if a occurs m times as a root in the auxihary equation, assxmie y = Ax"" ef". 3. If f (x) = a cos ;8x -|- 6 sin /3x, assume y = A cos fix + B sin. /3x, 62 DIFFERENTIAL EQUATIONS Chap. IV but, if COS fix and sin fix occur in the complementary func- tion, assume y = x[A cos fix + B sin fix]. 4. If / (x) = ae"^ cos fix + he"^ sin fix, assume y = A^" cos fix + Be"'' sin fix, but, if e"" cos fix and e"^ sin /3a; occur in the complementary- function, assume y = xe"" [A cos fix + B sin ;8a;]. If / (a;) contains terms of different types, take for y the sum of the corresponding expressions. Substitute the as- sumed value of y in the differential equation and determine the constants A, B, C, etc., so that the equation is satisfied. The general principle in the above rules is to express y as a linear function of all the distinct kinds of functions in S (x) and its derivatives of all orders. The exceptions to the various rules occur when some of the terms in the as- sumed value of y occur in the complementary fimction. Example 1. -7^^ + 4y = 2x + 3. A particular integral is evidently The solution of 2/=^(2x + 3). g+^» = ° IS y = Ci cos 2 a; + C2 sin 2 x. Hence the solution of the original equation is y = Ci cos 2 X + C2 sin 2 a; + J (2 X -h 3). Example 2. {D^ + SD + 2) y = 2 + s^ Chap. IV LINEAR E( ^UATIONi Assume y = A + Be\ Substituting this value for y 2 A + 6 Be^ = 2 + e= Hence 2A =2, 65 = 1 and A + 5e^ = !+■.. 63 Hence y = 1 + -e^ + cie-^+ c^e''^. The roots of the auxihary equation are 0, 0, — 1. Since is twice a root, we assume y = x^ {Ax^ + Bx + c). Substituting this value, l2Ax^+{24.A+QB)x + &B + 2c = x\ Consequently 12A = 1, 24:A + QB = 0, 6B + 2c = 0, whence ^ = 12' ^=-3' '=^- The solution is y = jxx^— ^x^ + x^ + ci + C2X + Cae-*. 32. Simultaneous Linear Equations. — We consider only linear equations with constant coefficients, containing one independent variable and as many dependent variables as 64 DIFFERENTIAL EQUATIONS Chap. IV equations. All but one of the dependent variables can be eliminated by a process analogous to that used in solving linear algebraic equations. The one remaining dependent variable is the solution of a Hnear equation. Its value can be found and the other dependent variables can then be determined by substituting in the preceding equations. If possible the work should be so arranged that after the first variable is found the others can be determined without integration. If integration is used in determining these later variables, the constants of integration may not all be arbitrary. It is then necessary to substitute the values found in the differential equations to determine the relations between the constants. Example. -^ — 3x — y — e' Using D for t; » these equations can be written (D^ _ 3) a; _ 2/ = e' -2x-\-Dy = Multiplying the first equation by 2 and the second by D^ — 3, we have 2 (D^ - 3) a; - 2 2/ = 2 e', - 2 (Z)2 _ 3) a; + (D2 - 3) Dy = 0. Adding, we get (D^ - 3 D - 2) 2/ = 2 e'. This equation, containing only one dependent variable, can be solved for y giving 2/ = (ci + c4) e-' + cae" — ^ e'. Substituting this value of y in the second equation, we find X = ^Dy = - (c2 - ci) - Cit e-' + cae"* - ^e'. Chap. IV LINEAR EQUATIONS 65 EXERCISES ' da? dx g ^_ 2^ -3^ = 0. tte* dx' dx - s- 10. 3 + y=2-.. 11. 3-42, =x^. <<. dy 13 ^ _ # = X. dx* dx 16. ^ - 0*2/ = e«^. ax' 16. jj + <»^ = •'OS "*• „. g-. = .»-!. "• 3-4j + 32/ = «^^='- 19. T^ - 9 2/ = e^ cos a;. 66 DIFFERENTIAL EQUATIONS Chap. IV 22. dx dt 2/ + 1, dy, dt = a; + l. 23. dx dt x-2y, dy _ dt X -y. 24. A dx -|+= X = . , dx «•''*' di + y 26. d?y dP d^x ' ^' dt^ = V- s^f^l^ = ^, dx d?y dt "•" dt^ = 1. cos t. 33. Vibrating Systems. — When a body is given a slight displacement from a position of stable equilibrium and re- leased, it vibrates about the position of equilibrium. If the resistance is neglected the differential equation of motion usually has the form §=+^^. = 0. The solution of this equation can be written a; = ^ cos {kt + 6), (33a) when A and d are constants. The motion is called har- monic. A is called the amplitude and the phase angle. In a complete vibration (across and back) the angle kt + e increases 360°, or 2x. The time of vibration is therefore k ' If the resistance is proportional to the velocity, the dif- ferential equation has the form d^x dx , df + ''di+^ = ^- The solution of this equation has the form X = Ae-^' cos (fit + e). (33b) Chap. IV LINEAR EQUATIONS 67 If at time t, the body is at the end of a swing, is the amplitude of that swing. The amplitude thus de- creases with the time. In some cases the differential equation has the form ^+k^sme = 0. This is not a linear equation. It becomes hnear, however, if we replace sin d by 6. For small angles this is a good approximation. Thus, when the angle is 5° sin e = .08716, e = .08727, showing that the error is less than 1 part in 800. Even for angles of 10° the error is only about | per cent. Example. A pendulum, consist- ing of a particle of mass m sup- ported by a string of length I, swings in a medium which resists with a force proportional to the velocity. Find the time of vibration. The torque about, the point of suspension due to the weight of m is — mg sin Cj the negative sign being used because when 6 is positive the torque tends to decrease d. The velocity of m is ,d0 Fig. 33. I dt The resistance is then ^^i- 68 DIFFERENTIAL EQUATIONS Chap. IV The torque due to this resistance is -kl- de , di ' ''' the negative sign being used since ■ . de . when -rr- IS dt positive the torque is negativt !. The moment of inertia of m about is niP. The equation j.d^e df = T is therefore d'e dt — mg I sine "'■ dt Replacing sin 6 by 0, this becomes d^.kddg df'^ mdt'^l The solution of this equation has the form e = Ae-2^cos|^: + e-i^) 11. j^=-(ea+ e"^ ), if the axes are properiy chosen. 12. 65.5 lbs. ., t / mg 13. Parabola. '='=■ '^"V mg + fcV. 14- I" ■ 23. < = y -In (6 + V35). 15. 77.6 ft. 2.15 sec. g^ 17. About 7 miles per second. 24. « = — : In (9 + 4 V5). 18. 116 hours. o" _ 19. About 42J min. 25. i =—^-g^ (17 + 4 V18). 20. s = i g (sin a — m cos a) &. ^ (2i6)^ r ,fi + e-fi 1- 3 *- 2 -' 26. Maximum height 48,500 ft., range 36.7 mi. mvo cos a ,, - -(\ 27. X = ^ (1 - e m ), 2/ = -p (kvo sm a + mg) (1 — e m , ) r- • x^,%!=l 29 ^'-^=1. 30. Helix. 21. 74 ANSWERS 32. The elUpse, ^ = (- - ~) cos e + -A- 33. r = 2 ro 34. r = a cos e. 1 + cos 9 36. jj ffi^ sin 0. 37. It slides imtil i = =— ^ and then rolls. ^'*- " 8m+3M ^^- "-2m+3Jf 40. The distance M moves in t seconds is 2m +4:M Pages 66, 66 1- y = ci + 01&. 4. y = Ci cos a; + C2 sin x. 2. 2/ = Ci&: + cae'^. 5. 2/ = ci + cje-* + cae^^:. 3. 2/ = (ci + C2x)e2^. 6. 2/ = cic^ + cje-^ + f 3 sin a; + C4 cos x. 7. y = e^ fci cos a; + ca sin x]. 8. 2/ = e-J^ [ci cos (i v'3T)^^sin (|V3 x).] 9. 2/ = (ci + Czx + Csa:^)e^. 10. 2/ = 2 — .t; + ci cos X + C2 sin x. 11. 2/ = Cie^a: + C2e-2l - i x2 - |. 12. 2/ = Cie^ — 3 (sin x + cos x). 13. 2/ = Ci + Ci&i: — i x^ — X. 4 1 14. y = cie-^ + C2e-«» "*" ' ^ ~ 9 "'" 15 *^*- 15. 2/ = CifiO^ + 626-0^ + jj— xeas;. 16. 2/ = { Ci + 2~ ) sin ax + C2 cos ax. 17. 2/ = ciex + e-ia; ^C2 cos ^-^ + c, sin 2^) - x' - & 18. 2/ = Ci6^ + C26'* — i e^* sin x. ANSWERS 75 19. y = Ci^ + c^r^x + ^ e'-t (6 sin x — cos x). 20. y = ci + cix ■{■ ci?? + c^e-^ + t^^ (4 cos 4 a; — sin 4 x).^ 21. y = (ci + C22; + i a;') e-^ + i e^. 22. 2/ = cie' + C26-' — 1, a; = cie' — C2e-' — 1. 23. 2/ = ci cos t + C2 sin i, a; = (ci + ca) cos i + (cz — Ci) sin 2. 24. X = Cie-' + C2e-*, 2/ = cie-' + 3 cje-'' + cos i. 25. X = cie' + cje-' + cs cos < + C4 sin t, y = Cie' + der-t — c, cos i — C4 sin t. 26. X = ci+Cit+cit' - iff +et 2/ = C4- (ci+2c3)«-i(c2-l)«2-|c3<'+^««-e«. 3. Pages 68-70 2V 3^' 5. 2.24 sec. 6. X = a cos ( V * )) where a is the amount the string is stretched ! weight and x i ilibrium. (4+2.) \/|. by one weight and x is measured from the point where it would hang in equilibrium. 8. ^ + .0924 ^ + 88.83 x = 0. aP at 9. 2.49 sec. 10. 2.\/±. Sg' ' V3ag ^CE Bt VaLC - C^B? — e 2X. 'sin t — • V4:LE-C'R' 2LC 12. i = , __ e- 22, sin i 13. Z ( —z 1 ) , where 6 = u ■s/hml. 14. r = ^{fl^-e^C). 76 ANSWERS 4 ll k A/ — , the origin being midway between the centers. 15. X = c cos t _ 16. An ellipse. 17. i = i(/„ + Eo\/l)e'^'' + i{lo - £o\/f )e-'^? INDEX The numbers refer to the pages AmpMtude, 66. Auxiliary equation, 56. Beams, deflection of, 37-39, 51. Bending moment, 38. Cable, equilibrium of, 39, 52. Change of variable, 26. Chemical reactions, 31. Complementary function, 61. Concentration, 10. Constants of integration, 4. independent, 55. number of, 55. Continuity, equation of, 10. Cooling, rate of, 15. Deflection of beams, 37. Density of sea water, 16. Derivative relations, 4. Determination of constants, 4. Differential relations, 7. Dissolving, rate of, 10, 13, 17, 18. Electric circuits, 31, 69. Equations, first order, 1-32. Mnear with constant coefficients, 56-64. n-th order, 55. second order, 33-54. Equation of continuity, 10. EquiMbrium of a cable, 39. Exact equations, 19-21. Flow of heat, 11, 18. Flow of water from an orifice, 9. Friction, 15, 52, 54. Growth, of bacteria, 32. of yeast, 14. Heat, flow of, 11, 18. Homogeneous equation, 26, 26. Homogeneous function, 25. Integrating factor, 21. Interest, continuously com- pounded, 8, 15. Linear equations, first order, 21- 23. with constant coefficients, 56-64. Motion, in a straight Mne, 40. in a plane, 43. of an electron, 44, 53. of a projectile, 53. of the center of gravity, 42. Orbit of a planet, 53. Order of a differential equation, 1. Orifice, flow from, 9, 16, 17. Particular integral, 61. Pendulum, motion of, 67. Phase angle, 66. Polar coordinates, 43. Pressure of air, 16. 77 78 INDEX Radium, rate of decomposition of, 6, 15, 32. Reflector, elliptic, 15. hyperbolic, 15. parabolic, 8. Rotation, of a liquid, 15. of a rigid body, 47, 48. Second 54. order equations, 33- Second order processes, 13. Separable equations, 1-18. Separation of the variables, 1. Singular solutions, 36, 56. Solution of a differential equation, 1. Ventilation, 18. Vibrating systems, 66. Ty