IFFERENTIAL 
 EQUATIONS 
 
 INEERING & 
 
 MAURUS
 
 is DUE 011
 
 BY 
 
 EDWARD J. MAURUS, M.S. 
 
 PROFESSOR OF MATHEMATICS IN NOTRE DAME UNIVERSITY 
 
 GINN AND COMPANY 
 
 BOSTON NEW YORK CHICAGO LONDON 
 ATLANTA DALLAS COLUMBUS SAN FRANCISCO
 
 COPYRIGHT, 1917, BY 
 
 EDWARD J. MAURU8 
 
 ALL RIGHTS RESERVED 
 117.10 
 
 gftt 
 
 GINN AND COMPANY PRO- 
 PRIETORS BOSTON U.S.A.
 
 Scionces 
 tibra/y 
 
 
 PREFACE 
 
 The aim of the author, in preparing this work, has been 
 to afford his classes an easy, condensed course in ordinary 
 differential equations, and to serve as a review of Integral 
 Calculus. 
 
 With few exceptions, the numerous problems are new, 
 though fashioned after the old models. 
 
 EDWARD J. MAURUS 
 NOTRE DAME, INDIANA 
 
 iii
 
 CONTENTS 
 
 CHAPTER PAGE 
 
 I. DEFINITIONS. DERIVATION OF A DIFFERENTIAL EQUA- 
 TION 1 
 
 II. DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND 
 
 FIRST DEGREE 4 
 
 III. DIFFERENTIAL EQUATIONS OF THE FIRST ORDER BUT 
 
 NOT OF THE FlRST DEGREE 14 
 
 IV. DIFFERENTIAL EQUATIONS OF ORDERS HIGHER THAN 
 
 THE FIRST * 19 
 
 ANSWERS 43
 
 AN ELEMENTARY COURSE IN 
 DIFFERENTIAL EQUATIONS 
 
 vii
 
 DIFFERENTIAL EQUATIONS 
 
 CHAPTER I 
 
 DEFINITIONS. DERIVATION OF A DIFFERENTIAL 
 EQUATION 
 
 Any equation containing differentials or derivatives 
 is called a differential equation. If only one independ- 
 ent variable is involved, the equation is an ordinary 
 differential equation. The following are examples of 
 ordinary differential equations: 
 
 (111 
 
 x - + y = xy cot x. (1) 
 
 e x dy + ye x dx = dy. (2) 
 
 The order of a differential equation is the order of 
 the highest derivative in the equation, for example, 
 (1), (2), and (3) are of the first order; (4) and (5) 
 are of the second order. 
 
 The degree of a differential equation is the degree of 
 the derivative of highest order in the equation, for 
 
 1
 
 2 DIFFERENTIAL EQUATIONS 
 
 example, (1), (2), and (4) are of the first degree ; (3) 
 and (5) are of the second degree. 
 
 A solution of a differential equation is an equation of 
 relationship between the variables, free from differen- 
 tials and derivatives, which will satisfy the differential 
 equation. In the process of solving, one or more inte- 
 grations must be performed, each involving an arbitrary 
 constant. The general, or complete, solution thus contains 
 an arbitrary constant for each unit in the order of the 
 equation ; that is, the complete solution of a differential 
 equation of the second order will contain two arbitrary 
 constants, etc. The differential equation may be con- 
 sidered as derived from its complete solution by the 
 elimination of the arbitrary constants. 
 
 Example 1. Form a differential equation by eliminating the 
 constant c from the equation xy = c sin x. Since two equations 
 are required to eliminate one arbitrary constant, a second equa- 
 tion is needed. This may be formed by differentiating the given 
 equation : xdy + ydx = c cos xdx. Eliminating the e, we have the 
 
 differential equation x + y = xy cot x. 
 ax 
 
 Example 2. Form a differential equation by eliminating the 
 constants c : and c 2 from the equation y = c 1 e r + c 2 e 2x . Since 
 there are two constants to eliminate, three equations are needed. 
 The two extra equations are formed by differentiating the given 
 equation twice, giving 
 
 ^ = c,e* + 2 c^ x and y - = c,e x + 4 c*e zx . 
 dx dx 2 
 
 Eliminating, by determinants or otherwise, we have 
 
 ^ 2 -3^ + 2, = 0. 
 dx 2 - dx
 
 DEFINITIONS 3 
 
 EXERCISE 1 
 Eliminate the constants from the following equations : 
 
 1. x 2 + f ex. G. y = c l e c + c z e 2x + c^ x . 
 
 2. my = m?x + 4:. 7. eV = c^e x + c^e~ x . 
 
 3. y 2 = 3 <?x + c 8 . 8. y = c x sin 8 x + c 2 cos 8 x. 
 
 4. y = c^ 2 * + c^ x . 9. y = c^ sec cc + c 2 tan cc.
 
 CHAPTER II 
 
 DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND 
 FIRST DEGREE 
 
 CASE I. When the variables can be separated, the 
 solution becomes a problem in direct integration. 
 
 Example. Solve the equation xdy ydx = dx dy, 
 Transposing and uniting, (x + 1) dy = (y + 1) dx. 
 
 Dividing by (x + 1) (y + 1), -^ = -^-- 
 
 y -\~ .1. x ~T~ X 
 
 The variables are now separated. 
 
 Integrating, log (y + 1) = log (x + 1) + c. 
 
 The c is an arbitrary constant of integration and may 
 be put in various forms suitable to the problem. The 
 simplicity of the solution generally depends upon the 
 selection of a proper form for the c. 
 
 In the example given, log c is as much an arbitrary 
 constant as c. Replacing the c by logc, the answer 
 assumes the form i 
 
 logO+l)=log(a;+l) + logc or #+l=<? 
 
 EXERCISE 2 
 Solve : 
 
 1. 2dx -f- xdy = 0. 3. cos* ydx + cos 2 xdy 0. 
 
 dit 4. tanxcos 2 ydx+ta,nycos*xdy=Q. 
 
 2. -f- + 2 xy = 0. 
 
 dx 5. xdy + ydx = 0. 
 
 4
 
 FIRST ORDER AND FIRST DEGREE 
 
 6. xydx + dy = x*dy. 
 
 7. tany Gos 2 ydx -f tan a; cos 2 xdy = 0. 
 
 8. e*dy + ye^efcc = dy. 
 
 9. sin a; cos ydx-\- cos ce sin 2/efa/ = 0. 
 
 10. cos 2 xdy -(- cos 2 ydx = cos 2 x cos*ydx. 
 
 11. a- 2 tZ?/ + sA&c + ?# + dy = 0. 
 
 12. xdy = vV +lcfce. 
 
 ay 
 
 13. a? -f 1 + y = xy cotx. 
 
 fKXf 
 
 14. yc?a; xdy = (x 2 + 1) % 
 
 15. x*ydy + xy^dx + sc<&e + yc?y = 0. 
 
 16. x 2 -7x + l2. + xy-5y-2 
 
 CASE II. If the equation is homogeneous in x and 
 #, substitute y = vz. Then separate the variables and 
 solve by Case I. 
 
 Example. 2 x z dy = 3 x 2 y dx y s dy. 
 
 Put y = vx; then rfy = vdx + xdv. 
 Substituting in the given equation, 
 
 2 vx 3 dx + 2 o; 4 <fo = 3 vx z dx v*x s dx v z x*dv. 
 Dividing by x s , 2 vdx + 2 xdv = 3 vdx v*dx v a xdv. 
 Separating the variables, 
 
 dx , t; 8 + 2 
 
 + -71 TT rft? = 
 # v (f 3 1) 
 
 ^r- - f-r- = 0. 
 
 w 1 v w^ + u + l
 
 DIFFERENTIAL EQUATIONS 
 
 Integrating, log x + log (v 1) 2 log v + log (v 2 + v + 1) = log c. 
 Uniting and dropping the logarithms, 
 zQ 8 -l) _ 
 
 Substituting - = t>, z 8 y 8 = cy 2 . 
 :r 
 
 EXERCISE 3 
 Solve : 
 
 2. (x + 5y}dx+(x+7y)dy = 0. 
 
 3. (2x 3y)dx + (4x 5y)dy = 0. 
 
 4. (x 2 + /) cfcc + zyefy = 0. 
 
 5. (x 1 y*)dx + xydy = 0. 
 
 6. (x* + 2xy + 3y*)dx+(x* + 6xy + 5y*)dy = 0. 
 
 7. xdx 4- iccfy + 2ydx = 0. 
 
 8. x^x + ydy = (x 2 + y^ydx. 
 
 9. 2 xytix (2 x 2 + t/ 2 )^ = 0. 
 10. 
 
 11. 
 
 12. 
 
 dy = _3Jy_ dy _ 
 
 " dx 3x* + * " dx 
 
 CASE III. If the equation is of the form 
 
 (OjZ + 6^ + c x ) e?a; + (a,^ + b$ + c 2 ) (y = 0, 
 
 it is called nonhomogeneous of the first degree. Substitute 
 x d + x and y = y' + y y Then dx = dx 1 and dy = dy\ 
 and the equation becomes 
 
 = OK
 
 FIRST ORDER AND FIRST DEGREE 7 
 
 % 
 
 Place a^ 4- \y Q + e 1 = and a z x + J^ + c 2 = 0. If these 
 equations are not identical or contradictory, they can 
 be solved for x and y^ These values, being substituted 
 in equation (a), reduce it to 
 
 O^' + ^y ) dx' + (a z x' + 6 2 y ) dy' = 0, 
 
 which is homogeneous, and solvable by Case II. 
 
 If the auxiliary equations are either identical or con- 
 tradictory, place a^x + b^y = v, eliminating x or y, and 
 solve the resulting equation. 
 
 EXERCISE 4 
 
 2. (x + 3y 2)dx + (5x + 7y-W)dy= 0. 
 
 3. (x 2 y + 3~)dx+ (3x - 6^ + 10)^ = 0. 
 
 4. (6x-8y + l)dx + (3x-12y + W)dy=Q 
 
 5. (5x + 62/- + 8)rfa; +(7x + 8y +10)dy = 0. 
 
 6. (cc + 2?/ + 3)<fo+(4a; + 5?/ + 6)^ = 0. 
 
 7. (3x-4:y-5)dx+(5x-6y-7)dy = Q. 
 
 8. (4x 5y + 6)dx-(5x + 3y ll)dy= 0. 
 
 ^ 
 ' dx 
 
 = 
 
 ' dx 
 
 CASE IV. A differential equation formed from its 
 primitive by differentiation without reduction is called 
 an exact equation. If it is of the first order, it can be 
 put in the form Mdx+Ndy = 0.
 
 8 DIFFERENTIAL EQUATIONS 
 
 Example 1. xeV + 2 x*y + 3 y cos x + y 3 = k. 
 Differentiating, we obtain 
 
 (e + 6 x*y - 3 y sinar)rfx + ( x e + 2 X s + 3 cos* + 3 y z )dy = 0. 
 Call the original expression u. Then M = e y + 6 x z y 3 y sin x, 
 
 which is > and N= xe" + 2x* + 3 cos a; + 3 y z , which is . 
 dx dy 
 
 Since - = -- 5 the test for an exact differential equation 
 
 of the first order is as follows : Obtain -- and --- If these 
 
 du dx 
 
 results are identical, the equation is exact. 
 
 To solve : Integrate M dx with respect to x. This will 
 give all the terms of the solution which contain x. If 
 the solution has any terms not containing ar, they will 
 not appear in Mdx, but will appear in Ndy, from which 
 they may be obtained. 
 
 Example 2. (2 ax + 2 hy)dx + (2 hx + 2 by)dy = 0. 
 
 Here M= 2ax + 2hy and N= 2 hx + 2 by. 
 
 SM 8N 
 
 = 2 h and - = 2 h. 
 
 dy dx 
 
 Hence the equation is exact. 
 
 Integrating the first part with respect to x, we have ax z + 2 hxy. 
 This gives all the terms involving x. The second part contains 
 one term, 2 by dy, without an x. Integrated, this gives by 2 . Hence 
 the complete solution is or 2 + 2 hxy + by 2 = c. 
 
 EXERCISE 5 
 
 1. (x-3y)dx-(3x-8y)dy = Q. 
 
 2. (7 x + 8 y} dx + (8 x + 9 y)dy = 0. 
 3. 
 
 4. (x 2 + 2xy)dx + x*dy = 0. 
 5.
 
 FIRST ORDER AND FIRST DEGREE 
 
 6. xy*dx -f- v?ydy + xdx + ydy = 0. 
 
 7. (4 x*y - 3 ccV) <fcc + (x 4 - 2 a; 8 ?/) dy = 0. 
 
 8. sina; cos ydx + cosx sinydy = 0. 
 
 9. (3arV-3e2'-2ze 3 2' + 2/ 8 )< c 
 
 + (a; 8 - 3xe y - 3x*e* 1J + 3xy*)dy = 0. 
 
 10. tany efce -f- x setfydy = 0. 
 
 11. sm~ l ydx + y = 0. 
 
 VI -/ 2 
 
 nydx = 
 
 " iy i+1 
 
 e y (e x + 2 e v ) dy = 0. 
 = e x dy e y dx + ye x dx xe v dy. 
 
 _ xdy 
 
 = 
 
 xy 
 
 CASE V. If the differential equation has been re- 
 duced after differentiation, it may no longer be exact. 
 
 Example. 2 x s y* - 3 x 5 y 2 = c. (1) 
 
 Differentiating, 
 
 (6 aty* - 15 zV) dx + (8 a; 8 /- 6 x 6 y) dy = 0. (2) 
 
 This is exact. Dividing by x z y, 
 
 (6 f - 15 x 2 y) dx + (8 zz/ 2 - 6 a; 3 ) dy = 0. (3) 
 
 Here M = 6 y s - 15 a; 2 y and JV = 8 a^ 2 - 6 ar 8 , 
 
 and = 18w 2 -15x 2 and = 8w 2 -18a: 2 . 
 
 dy Sx 
 
 Hence equation (3) is not exact. The factor x*y, which 
 would reduce equation (3) to (2) and make it exact, is 
 called an integrating factor. Sometimes the integrating
 
 10 DIFFERENTIAL EQUATIONS 
 
 factor for a given equation can be readily found by in- 
 spection. Various rules might be given for other cases. 
 The following will prove useful in many problems : 
 Multiply the given equation by x m y n . In the new 
 
 equation get - - and - If the method is possible, 
 dy dx 
 
 similar terms will be found. Equate the corresponding 
 coefficients and solve for m and n. 
 
 EXERCISE 6 
 
 1. (9x + ly)dx+7xdy = 0. 
 
 2. (x + 2y)dx + xdy = 0. 
 
 3. (12xy-15'y 2 )dx+(Wx 2 -18xy')dy = 0. 
 
 4. (16 xy - 15 y*)dx + (8ar 2 - 21 xy)dy = 0. 
 g dy _ 3 if 4 xy 
 
 dx x 2 2 xy 
 
 6. (5xy 2 if)dx + (5 a; 2 4 xy)dy = 0. 
 
 7. (35 xy + 20 if) dx + (18 x 2 + 33 xy}dy = 0. 
 
 8. (zV - y-)dx + (2x*y + x)dy = 0. 
 
 9. (2 xy* - y) dx +-(y + 2 a?) dy = 0. 
 
 xrfy 15 y*-35 x 2 
 
 10. ^ -\ -- * - = 
 
 ydx ^ 21y B -20x* 
 
 CASE VI. An equation in the form ~ + Py = Q, 
 
 dx 
 
 where P and Q are functions of x or constants, is called 
 linear. ef pdx is an integrating factor. 
 
 Example. - = xe x . HereP = -- ; \Pdx log:r = logar- 1 ; 
 fpdx dx x x J 
 
 e* x = x~ l . x~ l is the integrating factor. Multiply by this and 
 
 also by dx, and the equation becomes x~ l dy x~ 2 ydx e x dx. 
 This is exact. Integrating, x~^y e x + c, or y = x^e* + c).
 
 FIRST ORDER AND FIRST DEGREE 11 
 
 EXERCISE 7 
 
 dy y n dy 
 
 2. ~ - = oar. 4. -j- + y tan x = sec x. 
 
 6. cos x -^ + y = 1 sin x. 
 dx 
 
 (dy \ 
 
 6. cos x I ~ cos a; I = 1 w sin x. 
 \dx I 
 
 7. xdy -\- (xy -{- y^dx = dx. 
 
 8. oAfa/ -f xy(x + 2) 0*0; = 6*0*0;. 
 
 9. xc?y + 2/(?ic + dy e x (x 
 
 10 . ! + _^= = Va^pl-a, 
 
 nnp^ffll -J/ //-T* 
 * JU \Jvtl ^^ U lA/Ju 
 
 12. 
 
 CASE VII. If the equation is of the form 
 
 where P and Q are functions of x or constants, it can 
 be reduced to the linear form as follows : 
 Multiply by 
 
 tj 
 
 Substitute y~ n +1 = w, (n + \}y- n -^- = - The equa- 
 
 7 Cf3/ 0537 
 
 (T'W 
 
 tion becomes - +P( w +l)w= $( w+1), which is 
 dx 
 
 linear and can be solved by the preceding case.
 
 12 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 8 
 
 4. 3*|- y = * 
 
 5. i/dx xdy = ifefdx 
 
 y 
 
 2. 
 
 d 6. 3 
 
 3. 
 
 2 *ydx- 
 
 9. 4 y -r^ 3 y 3 tan x = 3 sec a;. 
 
 8. v 8 - 2 x 8 - 
 
 y dx 
 
 10. (2 x + 3)% = 4 y (2 x + 3) 2 </z + 
 
 11. x*e y dy = xdy ydx. 
 
 EXERCISE 9 
 MISCELLANEOUS PROBLEMS 
 
 ,. 
 
 3. (6* - 22 y + 22)<c + (17 x -6y-T)dy = 
 
 . = _ xe 
 
 dx x dx 
 
 6. 2y 2 dx = x(2y 
 
 7 . xdy = 2 ydx + x z e x dx. 
 
 8.
 
 FIRST ORDER AND FIRST DEGREE 13 
 
 . - 
 
 y dx dx 
 
 11. 4;X 
 
 12. 2x- = y 
 
 dx 
 
 13. sin 2 x dy = 2 tan y dx. 
 
 14. 
 
 15. (x 2y + 4)dx + (7x 142/ + l)dy = 0. 
 
 16. 2a; = 2- 
 
 72* 1 
 
 17. [ r- -- )c?ic ( = --- - \dy = 0. 
 
 \y 2 y/ \y 8 f) 
 
 18. (aa; + ^) dy = (by + a) dx. 
 19. 
 
 20. y (xy +l)dx + x (xy T)dy = 0. 
 
 21. 3afy% + ccy 8 ( + 2)rfcc 
 
 22. 2 tan t/cfcc + tan xdy = 0. 
 dy siny 
 
 dx cos x x 2 x cos y 
 
 24. (y 2 + 2?/ + 4)dx +(x 2 + 2x + 4)% = 0. 
 
 25. x (1 log xy) dy + y (1 + log xy) dx = 0. 
 
 26. (3- 
 
 27. 2sin 
 
 28. T/^ie = 2 x7/efa/ + 2 xy*dx. 
 
 29. 6xydy =(l-3y 2 + 5 
 
 30. xy(xdy + ye&c) + (aj^y y^c) Vl x 2 ?/ 2 = 0.
 
 CHAPTER III 
 
 DIFFERENTIAL EQUATIONS OF THE FIRST ORDER BUT 
 NOT OF THE FIRST DEGREE 
 
 When the equation is of the first order but not of 
 
 clu 
 the first degree, -j- will generally be represented by p. 
 
 QLjU 
 
 Three cases will be considered: (1) the equation solv- 
 able for p ; (2) the equation solvable for y ; (3) the 
 equation solvable for x. 
 
 CASE I. Solvable for p. If the equation can be 
 resolved into factors of the form 
 
 \j> -A (*> #)] O -/ 2 (* #)] etc -> 
 
 each factor may be put equal to and these equations 
 solved by the preceding chapter. 
 
 Example, p 2 2p 35 = 0. This can be factored into 
 O + 5)(/>-7)=0. 
 
 Placing each factor in turn equal to 0, and solving, we have 
 y + 5z + c = and y 7 a; + c = 0. 
 
 Since the differential equation is of the first order, the 
 solution can contain but one arbitrary constant. Hence 
 the c must be the same in each separate part of the 
 complete solution. This solution in the given example 
 may also be written (y -j- 5 x + c) (# 7 x -f <?) = 0. 
 
 14
 
 DIFFERENTIAL EQUATIONS OF FIRST ORDER 15 
 
 EXERCISE 10 
 
 1. p* - 6p + 8 = 0. 4. 2p 8 -3p*-3p + 2 = 0. 
 2.p 2 -p-l2 = Q. 5. p 3 - 9p 2 + 26p - 24 = 0. 
 
 3. p* p 2 12p = 0. 6. p 2 +px +py + xy = 0. 
 
 7. p 8 - 6p 2 x +11 px 2 - 6 x s = 0. 
 
 8. p\x+V)+y=p(x + y} + l. 
 9. x*p* -I=x 2 p-p. 11. jsV - y 2 = 0. 
 
 10. x 2 p 2 + a; 2 = ^? 2 . 12. xyp 2 -\-p (x 2 + y 2 ) + xy = 0. 
 
 CASE II. If the equation can be solved for y, do 
 so, and then differentiate with respect to x, remember- 
 
 ing that ~ = p. If possible, solve this equation by the 
 
 CL3C 
 
 second chapter. By eliminating the p between this 
 result and the original we obtain the required solution. 
 
 Example. p 2 x = 2py + x. 
 
 P 2 x-x 
 Solving for y, y = g- 
 
 Differentiating with respect to x, 
 
 r> = 
 
 Clearing of fractions and collecting terms, 
 
 z(p 2 + i~)-j-=p(p 2 + 1), or x-J-=p. 
 
 X 
 
 The solution for this equation is p = - 
 
 c 
 
 Eliminating the p between this and the original equation, 
 x 2 = 2 cy + c 2 .
 
 16 DIFFERENTIAL EQUATIONS 
 
 The equation y=px+f(p) is known as Clairaut's 
 equation. It comes under this case. Differentiating 
 
 with respect to x, we have p=p + &--+ f'(p)-r- or 
 j dx dx 
 
 rinf) 
 
 j D c +/'(j t O] = 0* Placing the first factor equal to 0, 
 
 dx dp 
 
 we have = 0, or p = c. Hence the solution of the 
 dx 
 
 differential equation is y = ex +/(<?). 
 
 If the second factor is placed equal to 0, and the p 
 eliminated between this and the differential equation, 
 the result will be a solution satisfying the differential 
 equation. It will contain no constant of integration, 
 and hence is not the complete solution. Neither can 
 it be obtained from the complete solution by giving 
 a particular value to the constant. It is called the 
 singular solution, and is, generally, the envelope of the 
 family of lines represented by the complete solution. 
 
 EXERCISE 11 
 
 1. y = 2px p*. 4. 4ar 2 r/ = 2px* -\- p*. 
 
 2. y = p + p 2 e~ 2x . 5. 6px 2 + 4ji> 2 = 9xy. 
 
 3. y =px logx +/> 2 ar 2 . 6. p*x = 3p^y + x. 
 
 The following are types of dairaufs equation. Solve 
 and obtain the singular solution if possible. 
 
 7. y =px +p 2 . 10. py = jj*x + 
 
 8. y =px + tan -1 p. 11. y = px + p Vp 2 + 1. 
 
 9. y=px + 6Vp 2 +l. 12. y 2 -l+p 2 (x*-l)=2pxy. 
 
 CASE III. If the equation is solvable for x, differ- 
 
 ed 1 
 
 entiate with respect to y. remembering that = 
 
 dy p
 
 DIFFERENTIAL EQUATIONS OF FIRST ORDER IT 
 
 Solve this equation, if possible, by the preceding chapter, 
 and eliminate the p between the resultant and the origi- 
 nal equation. 
 
 Example, y 3 px + p*y. 
 
 11 p*y 
 
 Solving tor x, x = - - - 
 
 3p 
 
 Differentiating with respect to y, 
 
 Clearing of fractions and collecting terms, 
 
 or =_2. 
 
 p y 
 
 Solving, p = L. 
 
 Eliminating the p by substituting in the original equation, 
 
 + c 8 . 
 
 EXERCISE 12 
 
 1. y = 2px + 4=p 2 y. 6. x y=p 8 .* 
 
 2. x-2p + 3p* = 0. 7. p*x=p + e-* y . 
 
 3. y = 2px + y* (2p)$. 8. 9 yp a 6 y^p + 4 = 0. 
 
 4. 2j9 8 a; - 3p*y = 2y. 9. jp&v =pe* v + e*. 
 
 5. j/e 2 ^ + 2px = 1. 10. 4 j p 2 xy 2 = 2_py" +1. 
 
 * When the elimination of p is impossible, the solution is indicated 
 by obtaining x and y in terms of p.
 
 18 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 13 
 MISCELLANEOUS PROBLEMS 
 
 1. jox 2 = 2 xy -+- x*p*. 
 
 2. e v =pe px . 
 
 3. z x= 
 4. 
 
 5. 
 
 6. p\x* - a 2 ) + if = 
 
 7. e 2 *=X*+^ 2ey - 
 
 8. 2pxy = xY + y*-a z (j?+ 1). 
 
 9 . 2 x 8 y = 3 x*p + 7/*p 8 . (Substitute y 3 = v and solve for v.~) 
 
 10. y 2 + 2/ 4 + a^P 8 = 2 zgp + 2 xtfp. 
 
 11. x(^ 2 + 2)=p(2ar l +l). 
 
 12. y(l 4 a;p 2 ) = 2^(05 ^ + 1). (Substitute T/ 2 = y and 
 solve for v.~) 
 
 13. 4^+130;^ + 92/y = 0. 
 
 14 . y 2 (p a + 1) = 4 (x + ypf. (Substitute 4 x 2 + 3 f = tr 2 .) 
 15. 4 cc + 9 jrp 2 = 6j9 (x + y). 
 
 17. V(* + y) 
 
 18. 2 aV + 5 xyji + 2 if = 0. 
 
 19. x*y + yp 2 = x*p. 
 
 20. >=e-*?- 
 
 
 21. j9 (x 2 - if) + xy (> 2 - 1) = a 2 /). Substitute
 
 CHAPTER IV 
 
 DIFFERENTIAL EQUATIONS OF ORDERS HIGHER THAN 
 THE FIRST 
 
 If the dependent variable and its derivatives are all 
 of the first degree and are not multiplied together, the 
 equation is called linear. Equations of this kind can all 
 be put into the general form 
 
 d n y _i_ A n ~ l - 
 
 A ~~ 
 
 J n1 j n 2 w-rit/ "vx' 
 
 aa; cte 
 
 where ^, A v A B , etc. are functions of x or con- 
 stants. We shall consider first the equation all of 
 whose coefficients are constants and whose second 
 member is 0. 
 
 To solve, substitute y = e kx . This will lead to an 
 auxiliary equation in Tc. Obtain the roots of this equa- 
 tion, & 1? & 2 , etc. The differential equation is satisfied by 
 any , of these values in y = e kx or by the combined form 
 y=c 1 e k * e +C2 i e' e * c -\- etc. The latter is the complete solution. 
 
 Example. - 2 8 y = 0. 
 
 dx 2 dx 
 
 Substitute y = e kx : then = ke kx and ^ = k z e kx . These, in 
 dx dx 2 
 
 the differential equation, give e kx (P-2 fc- 8) = 0, or & 2 -2 -8 = 0. 
 The roots of this equation are 4 and 2, and the complete solu- 
 tion is y = c l e ix + c 2 e~ 2x . 
 
 19,
 
 20 DIFFERENTIAL EQUATIONS 
 
 Three cases are to be considered : (1) when the roots 
 of the auxiliary equation are all real and different; 
 (2) when some or all of the roots are complex ; (3) when 
 some of the roots are repeated. 
 
 CASE I. This case is solved like the problem given. 
 
 EXERCISE 14 
 
 5. + 4 = . 
 dx 2 dx 
 
 dx 
 
 
 
 dx 
 
 CASE II. Some of the roots complex. The complex 
 roots may be written in the form a b \^T, and these 
 occur as conjugate pairs. The terms corresponding 
 would be
 
 ORDERS HIGHER THAN THE FIRST 21 
 
 The quantity in the parentheses may also be written 
 Cj cos bx + CjV 1 sin bx + c 2 cos bx c 2 V 1 sin bx, 
 or (cj + c 2 ) cos bx -f- (^V-- 1 c z V l) sin &c, 
 or, again, ^ cos fa + & 2 sin &c. 
 
 The terms then become e ax (^k 1 cos 5a; + & 2 sin 60;). 
 
 Example. - + 2 + 4v = 0. The auxiliary equation is 
 </z 2 ax 
 
 jfc 2 + 2 k + 4 = 0, of which the roots are 1 + V 3 and 
 1 V 3. Here a = 1 and b = v3, and the complete solu- 
 tion is y = e~ x (c 1 cos \/3 x + c 2 sin Vs x). 
 
 EXERCISE 15 
 
 c 
 
 If the operator be replaced by Z>, -j-^ by J5 2 , etc., 
 
 -, . CL3C* CL^u 
 
 the equation 
 
 ^-2^-8^ = 
 dx 2 dx 
 may be written 
 
 2Dy-8y = Q, or (Z> 2 - 2D - 8)y = 0.
 
 22 DIFFERENTIAL EQUATIONS 
 
 The coefficient of y is the same function of D as 
 the auxiliary equation is of k. If this coefficient is 
 factored as an ordinary quantic, the equation becomes 
 
 If the y is operated on by the near factor (-D+ 2), the 
 
 dt/ 
 
 result is -p- + 2 y ; operating on 
 ax 
 
 other factor (Z> 4), we obtain 
 
 dt/ 
 
 result is -p- + 2 y ; operating on this result with the 
 ax 
 
 the original differential expression. It can easily be 
 shown that the order of the factors may be changed 
 without altering the result. 
 
 CASE III. When the auxiliary equation has multiple 
 roots. 
 
 If two of the roots, k v k 2 , of the auxiliary equation 
 are equal, we cannot use c l ^ x + c^e***, as this could 
 be written (^ + c^)e k -p, and the '<?j + <? 2 is equivalent to 
 one constant. The way to handle this case can best be 
 illustrated by an example, 
 
 The roots of the auxiliary equation are 3, 3. Using the 
 D symbol, the differential equation may be written 
 
 (Z> 2 -6Z>+9)y=0, or (D- 3)(Z>- 3)y = 0. 
 Placing (Z> - 3) y = v, (1) 
 
 the equation becomes 
 
 (Z>-3> = 0, or ^-3^ = 0.
 
 ORDERS HIGHER THAN THE FIRST 23 
 
 This equation is linear with e~ Sx as the integrating 
 factor. Its solution is v = c 1 ^ x . Substituting this in (1), 
 
 we have -, 
 
 (D-^y = c^ x , or -/- 3 y= c, x . 
 ax 
 
 This equation is again linear with e~ Bx as the integrat- 
 ing factor. The solution is y = (c 1 x + c^e? 31 , which is 
 the complete solution of the original equation. In the 
 same way, if three of the roots of the auxiliary equa- 
 tion were equal, y = e k * x (c l y?+ c^x-\-c^) would be the 
 corresponding part of the complete solution, etc. 
 
 EXERCISE 16 
 
 -a 
 
 4. _ 4 + 4 
 dx* dx s dx 2 
 
 d^y d*y 
 ' ~ 5 ~- 
 
 7. 
 
 8. (Z) 3 + 5 Z^ + 3 D - 9)y = 0. 
 
 9. (T) 3 + D? - 5 D + 3) y = 0. 
 10. (Z) 3 - 6Z> 2 +12Z)- 8)y = 0.
 
 24 DIFFERENTIAL EQUATIONS 
 
 CASE IV. When the second member is not 0, the 
 complete solution will consist of two parts : the general 
 part, which contains the constants of integration ; and the 
 particular part, which accounts for the second member. 
 The complete solution is the sum of the two parts. The 
 general part is obtained by one of the three preceding 
 cases ; the particular part depends upon the second 
 member, and various methods might be given for ob- 
 taining it. We shall confine ourselves to a second 
 member comprising terms of the forms of 1 , e nx , sinnx, 
 cosnx, and their products, and shall use an inspection 
 method to arrive at a corresponding solution. 
 
 Example 1. | ^ 6w = 3z 2 oz + 6. Here the auxiliary 
 
 dx 2 dx 
 
 equation is fc 2 k 6 = 0, its roots are 3 and 2, and the general 
 part of the solution is y = c 1 e sx + c z e~ zx . The second member of 
 the given equation must have come from a power series of the 
 form c s x z + c^x + c 6 . Substituting y = c s x 2 + c t x + c 5 in the given 
 equation, we have 
 
 2 e g 2 c s x c 4 6 c 3 x 2 6 cx 6 c 5 = 3 x 2 5 x + 6. 
 Equating the coefficients of like powers of x, 
 
 -6c 8 = 3; -2c 3 -6c 4 =-5; 2 c 3 - c 4 - 6 c 6 - 6. 
 Solving these equations, 
 
 c s = 2 5 C 4 1 j C 5 = 3- 
 
 x z 4 
 Hence the particular part of the solution is \- x > and 
 
 x 2 4 
 
 the complete solution is y = c^ x + c z e~ 2x h x 
 
 2 3 
 
 d u dy 
 
 Example 2. - 5 + 4w = 2e 8;r . Here the auxiliary equa- 
 rfx 2 dx 
 
 tion is k z 5 k + 4 = and the general part of the solution is 
 y = c 1 e tx + c^. The second member of the given equation, 2 e sx ,
 
 ORDERS HIGHER THAN THE FIRST 25 
 
 must have come from a term of the form ce ax . Substituting 
 y = ce sx in the given equation, 
 
 9 ce ax - 15 ce 8x + 4 ce 8x - 2 e ax , or - 2 c = 2. 
 .-. c=-l. 
 
 The particular part of the solution is, then, e 8x , and the 
 complete solution is y = c l e* x + c 2 e x e sx . 
 
 ff^jf dy 
 
 Example 3. ^ 5 -j- + 6 y = sin x. Here the general part of 
 
 the solution is y = c^e ix + c 2 e sx . The second member, sin x, must 
 have come from an expression of the form c 3 sin x + c t cos x. 
 Substituting y = c 3 sin x + c 4 cos x in the given equation, 
 
 (5 c 3 + 5 c 4 ) sin x + (5 c 4 5 c 3 ) cos x = sin z. 
 Equating corresponding coefficients, 
 
 5 c 3 + 5 c 4 = 1 ; 5 c 4 5 c 3 = 0. 
 Solving these equations, c 3 = c 4 = ^ 
 
 Hence the complete solution is 
 
 sin a: + cos a; 
 ^ = c^ x + c 2 e 8X + -- --- 
 
 Example 4. + 2 = e~ zx . Here the general part of the 
 rfx 2 dx 
 
 solution is y = c x + e 2 e~ 2a; . It might appear that the second 
 member must have come from a term of the form ce~ 2x ; but 
 this is already found in the general part of the solution. In this 
 case assume y = cxe~ 2x . Substituting in the given equation, 
 
 4 cxe~* x - 4 ce~ zx + 2 ce~ 2x - 4 cxe~ 2x = e~ 2x . 
 
 .: -2c 3 = l, or c 3 =- $. 
 Hence the complete solution is 
 
 If the second member consists of several terms, the 
 corresponding particular integral may be found for the 
 parts taken separately.
 
 26 ' DIFFERENTIAL EQUATIONS 
 
 EXERCISE 17 
 
 2. 
 3. 
 
 4. -r^r 16 ?/ = sin x. 
 dx* 
 
 6. (T) 2 - V2.D + 27)y = 81 x - 171. 
 
 7. (D 2 -l)y = 3^- 
 
 8. ( J D 2 + 4)?/ = cos2cc. 
 
 9. (Z) 2 D 12) y = e 4a: + sin3a; + e 2a: sin3x. 
 10. 
 
 CASE V. When the equation is of the form 
 
 AlX n^ +A ^d^ ji _ 
 
 dx n dx n ~ l 
 
 A v A v etc. being constants, substitute x = e f . Then 
 
 ^ _ dy dt _ dt 
 dx dt dx e* 
 
 dy dy 
 
 or x -^- = -^-- 
 
 dx dt 
 
 Similarly, = - 
 
 J ' 2
 
 ORDERS HIGHER THAN THE FIRST 27 
 
 Using the symbol D for > we have 
 
 Ct>i> 
 
 x^ = Dy; 
 
 dx 
 
 rf^LJL _ (D3 _ 3z>2_|_ 2Z>)y 
 
 = D (Z> 1) (D 2)#, etc. 
 
 This substitution will reduce the equation to one of the 
 preceding forms. In the same way, if the equation is of 
 
 the form A^ax+by 1 -+^4 2 (#a;+5) n ~ 1 f- + . . . = 0, 
 
 dx n dx n 
 
 substitute ax + b = e*. 
 
 EXERCISE 18 
 
 cc (Zee 
 
 ^ 
 
 dt*~ t*
 
 28 DIFFERENTIAL EQUATIONS 
 
 8. (>x + 6) 2 -12aV = 0- 
 
 ' dy? 
 
 9 . 2*-3> 
 
 CASE VI. If the equation does not contain y directly, 
 represent the derivative of lowest order by p. Then the 
 
 next derivative will be -f-t etc. Solve this new equation 
 ax 
 
 for p in terms of x, and ultimately for y. 
 
 Putting &=p, X 4= p +p*. 
 
 ax ax 
 
 Separating the variables, 
 
 dp _ dx 
 p 3 + p x 
 
 dp p dp dx 
 
 or * . = 
 
 p p*+l x 
 
 . dy dif x 
 
 Solving for p = -f-, -f-= ; 
 
 dx dx Vc 2 x 2 
 
 whence y = V c\ x 2 + c 2 . 
 Clearing of radicals, 
 
 EXERCISE 19 
 
 ' --
 
 ORDERS HIGHER THAN THE FIRST 29 
 
 dy 
 
 j . 
 
 1 
 
 I 
 
 W l , o I I IT I * 7 
 
 ^aa;/ aa; 
 
 8. | = (2 tan x -j- cot a?) -^ 
 
 10. 
 
 CASE VII. If the equation does not contain x directly, 
 
 ^ = p; then ^l = ^P , ^l = p ^P. 
 dx dx 2 dy dx dy 
 
 Example. (1 - f) g + , (ff= 0. 
 
 Putting ^ = /, (1 - /)^ ^-+p*y = 0. 
 ax ay 
 
 Separating the variables, 
 
 dp _ ydy 
 
 ~' 
 
 / - 62/ 
 
 Solving, ^ = Cj V 1 - y* - -^ ; 
 
 whence
 
 SO DIFFERENTIAL EQUATIONS 
 
 EXERCISE 20 
 
 CASE VIII. When the equation is of the form 
 
 -^ =/(?/), multiply by 2 -~ dx and integrate. This ' 
 
 /dy\ z r 
 gives ( -j- \ = 2 I /(#) dy + <?. Extract the square root, 
 
 separate the variables, and solve. 
 
 d 2 y 
 Example. ^ = y. 
 
 f/y** 
 
 , n dy n dy d 2 y n dy 
 Multiplying by 2 > 2 ^ = 2w-^. 
 dx dx dx* dx 
 
 Integrating, CT = y*+ c x , 
 
 or - = Vf + Cl ;
 
 ORDERS HIGHER THAN THE FIRST 31 
 
 Using the positive sign, 
 
 i) = * + log c 2 , 
 y + 
 Clearing of radicals, y* + c^ y z 2 c 2 ye x + c|e aa: . 
 
 c e x c 
 Solving for y, 
 
 or # = A^ 
 
 EXERCISE 21 
 
 dx" dx 2 dx z 
 
 2. T^= V 6 -;r^ = ~/=' 8 ' y^T?^ 1 - 
 
 ttiC Ct/iXs "\/ -7/ {Zi(y 
 
 3- ^3^=1- 6. ^T^ = l. 9. ^v^=l. 
 
 cLy {iOi ci^f 
 
 , 
 
 10. cos 15 ?/ -7-^ = sin y. 
 
 tt*// 
 
 CASE IX. An exact differential equation has been 
 defined as one obtained from its primitive by differen- 
 tiation without further reduction. If the equation is 
 of an order higher than the first, and is of the form 
 
 V A< etc " 
 
 are functions of #, the test for exactness is as follows: 
 Differentiate A 1 with respect to x and subtract the 
 result from A 2 ; differentiate this remainder with respect 
 to x and subtract from A y etc. If the last remainder 
 is 0, the equation is exact. If the test is satisfied, the
 
 32 DIFFERENTIAL EQUATIONS 
 
 first integral is obtained as follows : The first term will 
 
 be A 1 ^ ; the coefficient of the second term will be 
 
 the first remainder obtained in the test ; the coefficient 
 of the next term the next remainder, and so on. The 
 
 second member will be / f(x~)dx + c. If the resulting 
 
 equation is again exact, repeat the process. Otherwise, 
 solve, if possible, by one of the preceding cases. 
 
 fi II fl tl 
 
 Example. (3 x 2 - 5 a;) - + (12 x - 10) ^ + 6 y = 
 
 'rfz 2 v dx 
 
 -,-,,,.,, ox o o 
 
 Following the test, 
 
 X 
 
 Since the last remainder is 0, the test is satisfied. The first 
 integral is , 
 
 (3 x 2 - 5 x) + (6 x - 5) y = c x 
 Qx-5 
 
 
 The test is again satisfied, and the complete solution is 
 (3 x 2 5 x) y = c x x + c 2 . 
 
 EXERCISE 22 
 
 2. (2 a5 + 5) 2 
 
 , v 
 
 3. sm x -r-^ + 2 cos x -r- y sm x = e 2 . 
 dor dx 
 
 4.
 
 ORDERS HIGHER THAN THE FIRST 33 
 
 6. 
 
 7. x 1 + 21 x 6 +126 o; 6 -f 210a;V = 3 e* + sin a. 
 ete 8 da; 2 <&c 
 
 9 . + = e _, 
 
 /7/>8 /V'*' 2 
 it-**/ u-x 
 
 10. (a 2 - z 2 )^ = a*. 
 7 cfo; 2 
 
 . 
 
 CASE X. In Case V a change of the independent 
 variable reduced the equation to a solvable form. In 
 many other problems a proper change of either the 
 dependent or independent variable will effect a reduc- 
 tion of the given equation. This is especially true of 
 equations of the second order. We shall consider two 
 cases, and shall illustrate the methods by examples. 
 
 d z ii J dti 
 
 Example. - -^ 2 tan x y = 0. 
 ax* . ax 
 
 dy dv du d z y d z v _ du dv d 2 u 
 
 Putting y = uv, -f- = w + v , = U- + 2- --- - + v 
 
 dx dx dx dx* dx 2 dx dx dx z 
 
 in the given equation and collecting terms, 
 
 d z v /.. du n . \ dv fd z u ~ , du \ 
 u - (2 -- 2 tana:) -- hi -- 2tanw u)v = 0. (1) 
 dx z \ dx /dx \dx z dx / 
 
 Placing the coefficient of equal to 0, and solving without 
 
 dx 
 constants of integration, 
 
 u = sec or.
 
 34 DIFFERENTIAL EQUATIONS 
 
 Then = sec x tan x, 
 
 dx 
 
 and - = sec 8 a: + sec x tan' 2 x. 
 
 dx 2 
 
 Substituting in equation (1) and simplifying, 
 
 whence v = c^x + c 2 . 
 
 Therefore y = uv = (c^x + c 2 ) sec x. 
 
 EXERCISE 23 
 + y(2 cot a z + 5) = 0. 
 
 3. ^ - (1 + 2 cota;)^ + y(l + cotx + 2 cotV) = 0. 
 
 4. 
 5. 
 
 6. ^ + 2 a cot ax ^ - 2 a 2 ?/ = 
 
 7. 
 
 8. 
 
 /W 2 */ \ /7 
 
 10.
 
 ORDERS HIGHER THAN THE FIRST 35 
 
 CASE XL In this case the equation is reduced by 
 changing the independent variable. 
 
 Example. 9 x f + 6 -*- + 4 x~ % = 0. 
 
 dx 2 dx / 
 
 Putting x =/(), 
 
 then ? = ^'' 
 
 dx dt dx 
 
 -* '& ^ 
 
 d z y d 2 y/dt\ z dy d 2 t 
 and = ( \ + ~ ' owr 0L\/ 
 
 Substituting in the given equation and collecting terms, 
 
 i/ 
 Placing the coefficient of equal to and solving, 
 
 t = 3 zi 
 Substituting in (1) and reducing, 
 
 ,fi + 4,=a 
 
 The solution of this equation is 
 
 y = CJL cos f < + c 2 sin <. 
 Hence y = c i cos 2 x^ + c 2 sin 2 x. 
 
 This result may also be written 
 
 y <?j sin (2 X s + ) 
 
 Placing the coefficient of -r*j equal to the coefficient 
 
 CvC 
 
 of y in (1) will often effect an easy reduction.
 
 36 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 24 
 
 . - 
 dx* dx 
 
 d*y 2x dy y = 
 
 " ^ ~ 
 
 4. - 
 
 dx 3 dx 
 
 6. 
 
 6 ' 
 
 7. 
 
 8 . 
 
 dx 
 
 d*y dy 
 
 9. -7* tan x -2- y sec-ic = 0. 
 dx 2 dx 
 
 10 . ^- 
 
 dx* dx 
 
 EXERCISE 25 
 MISCELLANEOUS PROBLEMS 
 
 3. (D 4 - 
 6. - 
 
 (Substitute s = log sec x.)
 
 ORDERS HIGHER THAN THE FIRST 37 
 
 7. 
 
 11. ( 
 
 12. 
 
 (Substitute y = ve*. 
 
 13. Z> 8 - 
 
 15. H +(tanx - 2 cotx) ^ + 2y cot 2 x = 0. 
 
 (Substitute z = log sin x.) 
 
 dy d*y 
 
 16. - = 35 
 dx dx 2 
 
 fl^ti (ill 
 
 17. (x log x) 2 -73 + x log x (log x 4)-T- + 6y = 0. 
 
 dx dx 
 
 (Substitute z = log log x.) 
 
 18. (> + 4)?/ = 0. 
 
 19. (T) 2 
 
 22 ~-a*s 
 22. - a. 
 
 23. Z> 4 - 
 
 1-logy. 
 logy
 
 38 DIFFERENTIAL EQUATIONS 
 
 dy 
 = 
 
 26. () 8 - 
 
 W.fi-2g 
 
 dx* dx 
 
 29. 
 30. 
 
 EXERCISE 26 
 APPLICATIONS 
 
 1. Find the equation of the curve whose subnormal is 
 constant. 
 
 2. Find the equation of the curve whose subnormal is 
 proportional to the square of the abscissa of the point of 
 contact. 
 
 3. Find the equation of the curve whose subnormal is 
 equal to the abscissa of the point of contact. 
 
 4. Find the equation of the curve in which the slope at 
 any point varies directly as the abscissa of the point. 
 
 6. Find the equation of the curve in which the slope at 
 any point varies inversely as the ordinate of the point. 
 
 6. Find the equation of the curve in which the slope at 
 any point varies directly as the ordinate of the point. 
 
 7. Find the equation of the curve whose slope at any 
 
 4a 
 point is TT
 
 APPLICATIONS 39 
 
 8. A point moves in a path always perpendicular to the 
 line joining its position to the origin. Find the equation of 
 its path. 
 
 9. The tangent to a curve cuts intercepts from the 
 coordinate axes whose sum is constant. Find the equation 
 of the curve. (Singular solution.) 
 
 10. The tangent to a curve cuts intercepts from the 
 coordinate axes whose product is constant. Find the equa- 
 tion of the curve. (Singular solution.) 
 
 11. Find the equation of the family of curves all of which 
 cut the hyperbola x 2 y 2 a? at right angles. 
 
 12. Find the equation of the family of lines all of which 
 cut the circle x 2 + y 2 = 2 at right angles. 
 
 13. A point moves so that its acceleration varies inversely 
 as the cube of its distance from the initial point. Find the 
 equation of its motion. 
 
 14. A point moves so that its acceleration varies inversely 
 as the square of its distance from the point of starting. Find 
 the equation of its motion. 
 
 15. A point moves so that its acceleration varies directly 
 as the distance from the initial point and is negative. Find 
 the equation of its motion. 
 
 16. Find the equation of the curve whose radius of curva- 
 ture is constant. 
 
 17. Find the equation of the curve whose radius of curva- 
 ture is twice the length of the normal. 
 
 18. Find the equation of the curve in which the radius 
 of curvature at any point varies directly as the slope at 
 the point.
 
 40 DIFFERENTIAL EQUATIONS 
 
 19. If a horizontal beam is supported at both ends and 
 a load is uniformly distributed along its length, the upper 
 fibers are in compression and the lower in tension. Between 
 the two is a neutral surface, and its intersection with a 
 vertical plane, parallel to the axis of the beam, is called the 
 neutral line. Its differential equation is 
 
 "(' A 
 
 da? 2 \4 / 
 
 the origin being at the middle of the beam, x horizontal, 
 and y vertical, I the length of the beam, w the weight per 
 unit of length, E the modulus of elasticity of the beam, / the 
 moment of inertia. Find the equation of the neutral line, if 
 
 ~ = when x = 0, and y = when x = - 
 dx 2 
 
 20. If a load P is concentrated at the center of the 
 beam and the weight of the beam neglected, the differential 
 equation of the neutral line is 
 
 Solve the equation under the same conditions as above. 
 
 21. A cantilever beam has one end unsupported. When 
 uniformly loaded, the differential equation of the neutral 
 line is 
 
 the origin being at the fixed end. Solve the equation under 
 
 u/U 
 
 the conditions, when x = 0, y = 0, -^ = 0.
 
 APPLICATIONS 41 
 
 22. If the weight P is concentrated at the free end and 
 the weight of the beam neglected, the differential equation 
 of the neutral line is 
 
 Solve under the same conditions as above. 
 
 23. If R is the resistance of an electrical current, C is 
 the current, V the electromotive force, L the coefficient of 
 
 dC 
 
 self-induction, L \- RC = V. Find C in terms of t. 
 
 (Lif 
 
 24. The equation for the current C in a circuit consist- 
 ing of a charged condenser and a resistance R is 
 
 dC C 
 
 where K is the capacity of the condenser. Find C in 
 terms of t. 
 
 25. The equation for the quantity of electricity dis- 
 charged by a condenser in a circuit consisting of a condenser 
 
 of capacity K and a resistance R is -f + r~= = 
 -. T . dt RK R 
 
 Find in terms of t.
 
 ANSWERS 
 
 EXERCISE 1 
 
 1. x 2 == ce-v. 
 Z.y = ce-^. 
 8. tan x + tan y = c. 
 
 10. tanx + tany = x + c. 
 
 11. x + y = c(l - xy). 
 
 12. vV + 1 = ex - y. 
 
 13. xy = c sin x. 
 
 EXERCISE 2 
 
 4. tan 2 x + tan 2 y = c. 7. tan* tan y = c. 
 5.xy = c. _ 8. y(eP c 1) = c. 
 
 6. y = c Vx 2 1. 9. sec x sec y = c. 
 
 14. T/ = c (x 2 + x + 1). 
 
 15. (x 2 + 1) (y 2 + 1) = c. 
 
 16. (y - 2) (x - 3) 2 = c (x - 4). 
 
 17. x?/ = c sin y. 
 
 2x) 2 (?/ 
 
 2. 
 
 3. log c (5 y 2 xy 2 x 2 ) 
 
 EXERCISE 3 
 
 4. x 2 (x 2 
 
 ViT 
 
 _ 
 
 5. x = ce 2a;2 . 
 6. 
 7. 
 
 43
 
 44 
 
 DIFFERENTIAL EQUATIONS 
 
 9. y = 
 
 10. 2x 8 + 9x2y + 9xy 2 
 
 11. x 2 + y 2 = ex. 
 
 12. x 3 + y s = ex 2 . 
 
 xf 
 
 13. y = ce" 3 . 
 
 14. x s y s cxy. 
 
 EXERCISE 4 
 
 2. (x + ly - 2) 2 (x + y - 2) = c. 
 8. 5x + 15y + c = log (5 x 
 
 5. (x + y + 1) (5x + Sy + 14) 2 = e. 
 
 7. log c (6 y 2 xy 3 x 2 + 23 y 8 x 
 
 g 12 y x 
 
 log 
 
 \/73 
 
 8. 4x 2 - 
 
 9. 7x 2 18xy 2y 2 + I4x-I8y 
 10. x 7y + c = log(3x by + 11). 
 
 1)V73 
 
 EXERCISE 5 
 
 1. 2x 2 3xy + 4 y 2 = c. 
 
 2. 7x a + 16xy + 9j/ 2 = c. 
 8. x 2 + Sxy + 7y 2 = c. 
 
 4. x 8 + 3x 2 y = c. 
 
 5. 2x 8 + 9x 2 y + 9xy 2 + 7 
 
 6. xV + x 2 + y 2 = c. 
 
 7. x*y - x 8 ^ 2 = c. 
 
 8. cosx cosy = c. 
 
 9. x z y 3xe+ x 2 e? + xy* 
 
 10. xtany = c. 
 
 11. xsin- 1 j/ = c. 
 
 12. y = c (x + l) n . 
 
 13. e 2 * + e^ + w + e^v = c . 
 
 14. 2x 2 y = j/e* xe^ + c. 
 
 15. Vx 2 + 2/ 2 = x + c. 
 
 16. x 2 y 2 = cxy. 
 
 EXERCISE 6 
 
 1. 3x 8 + 7x 2 y= c. 
 
 2. x 8 + 3x 2 y = c. 
 
 8. 2xV-3x 5 y = c. 
 
 4. x^(2 x 3 y) = c. 
 
 5, xy (x - y) = c. 
 
 6. 6xV - 3xV = c. 
 
 7. x ^l + x l y V- = c . 
 
 g. xV + y = ex. 
 9. x 2 y 8 -x-cy 2 . 
 10. 3 x 5 y 7 - 5 x V = c.
 
 ANSWERS 
 
 45 
 
 EXERCISE 7 
 
 1. xy = x s + c. 
 
 2. y = x(x 8 + c). 
 
 3. x s y = x + c. 
 
 4. y = sinx + c cosx. 
 
 5. ?/ (sec x + tan x) = x + c. 
 
 6. y = sin x + (x + c) cos x. 
 
 10. y (x + Vx 2 + 1) = x + c. 
 
 11. y = (x + 0)6*" -Jx. 
 12. 
 
 1. y~ = x(c 2x). 
 
 2. y- 3 = x(x + c). 
 
 3. x 2 y 3 = x 5 + 'c. 
 
 4. 22/- 3 = x- 1 
 
 5. x = ?/(e a: + c). 
 
 6. X s + y s = ex 2 . 
 
 EXERCISE 8 
 
 8. x 8 + y* = ex. 
 
 9. y^ = (x + c) sec x. 
 (4e*+3x). 10 2 ^(2x+3) 2 = c( 
 
 11. y = x(&> + c). 
 
 EXERCISE 9 
 
 1. y = clogxy. 
 
 3. 
 
 4. 
 
 5. y = x (ev + c) 
 
 6. y 2 = x 2 (e*' + c). 
 
 7. y = x 2 (ef c + c). 
 
 8. &> + 1 = c (e* + 1) 
 
 9. e*' = e* + c. 
 
 10. y = ce^. 
 
 11. 2x 2 3xy + 5y 2 
 
 12. x = ce^*. 
 
 13. siny = c tanx. 
 14. 
 
 15. 
 
 2. x = clogxy. 
 
 2 x 
 16. sin- 1 = x + c. 
 
 18. 
 19. 
 
 20. 
 21. 
 22. 
 23. 
 24. 
 25. 
 26. 
 
 27. 
 
 28. 
 29. 
 
 30. 
 
 sin - = ex 2 . 
 x 
 
 b) b 
 
 ce~ x . 
 sin 2 x siny = c. 
 x siny + x 2 y y cosx = c. 
 X + y + 2 = c(xy + x + i/ 2). 
 xlogxy -cy. 
 x 3 x 5 ?/ 2 = c. 
 
 sin- 1 - = ex 2 . 
 x 
 
 x = y 2 (x 2 + c). 
 x 3x?/ 2 + x 5 = c. 
 
 v 
 
 log- = c. 
 x
 
 46 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 10 
 
 1. (y- 
 
 2. (y - 4x + c) (y + 3x + c) = 0. 
 
 3- (y - c) (y - 4x + c) (y + 3x + c) = 0. 
 
 4. (y-2x + c)(i/ + x + c)(2y-x + c) = 0. 
 
 5. (2/-2x + c)0/-3x + c)(?/-4x + c) = 0. 
 
 6. (y - ce-x) (x 2 + 2 T/ + c) = 0. 
 
 7. (x 2 - 2 y + c) (x 2 - y + c) (3x 2 - 2 y + c) = 0. 
 
 8. y x + c; y l = c(x + l). 
 
 9. (y x + c) (xy ex 1) = 0. 
 
 10. x 2 + (y - c) 2 = 1. 
 
 11. (xy c)(y cx) = 0. 
 
 12. (xy - c) (x 2 + y 2 - c) = 0. 
 
 EXERCISE 11 
 
 2i *Q C O C X o Cy ~r 1. 
 
 I / *... .i, 
 
 3 p2 ' 7. y = ex + c 2 ; singular solution, x 2 + 4 ?/ = 0. 
 
 p2 2 c 8. y = ex + tan -1 c ; no singular solution. 
 
 3 9. y = 
 
 2. y = ce 1 + c 2 . singular solution, x 2 + y 2 = 36. 
 
 3. y = c log x + c 2 . 10. y = ex + - ; singular solution, y 2 = 4 ox. 
 
 4. y = ex 2 + c 2 . c 
 
 5. ex 8 = (y c) 2 . 11. y = ex + c Vc 2 + 1. 
 
 12. y = ex + Vc 2 + 1 ; singular solution, x 2 + j/ 2 = 1. 
 
 EXERCISE 12 
 
 1. y 2 = ex + c 2 . 3. y 2 = ex + cf 
 
 2. y =p 2 -2p 8 + e; 4. 27 cy 2 = 8(x - c) 8 
 x = 2p 3p 2 . 5. e 2 ? = 2 ex + c 2 . 
 
 3o 2 
 8 
 
 
 
 x = c - -f - Sp - 3 log(p - 1). 
 8 
 
 7. x = cev + c 2 . 9. e* = ce + c 2 . 
 
 8. cy 8 = (x + c) 2 . 10. x = cy* + c 2 .
 
 ANSWERS 47 
 
 EXERCISE 13 
 
 12. / 2 = 
 
 2. ?/ = cx + logc. c + 1 
 
 3. x = clogy + c*. 13. (4x 2 + 92/2-c)(x 2 + 2/ 2 -c) = 0. 
 
 4. 2/ = cx2aVc~. 14. (x + c) 2 + y 2 = 4 c 2 . 
 
 5. (x 2 - y 2 + c)(y - ex) = 0. 15 - (2x-32/+c)(2x 2 -3^ 2 +c)=0. 
 
 6. y = ex Voc a + 6 2 . 16 > * = cx ~ & + c& x - 
 
 7. e* = ce" + c 2 . 17. ( X2/ - c )(y - ce) = 0. 
 
 8. y = cx a Vc 2 + 1. 18. (xy 2 - c)(x 2 y - c) = 0. 
 
 9. 2 y 3 - 3 cx 2 + c 3 . 19. y 2 + c 2 = cx 2 . 
 
 10. y 2 = 2 cxj/ 2 + c^ 2 . 20. y 2 e* x + 2 cy + c 2 = 0. 
 
 11. d/-x 2 + c)(2/-logx + c) = 0. a = , a% 
 
 EXERCISE 14 
 
 1. y = c 1 e^ x + c 2 e-*. 
 
 2. y = c^^ + c 2 e 2 *. 6. y = c^ 21 + CgC 2 . 
 
 3. ?/ = c^x + c 2 e~ & *. 7. y = c x e 2a; + c z <? x + cjfi*. 
 
 4. ?/ = Cje 5 * + c 2 e- 4:r . 8. y = c^ x + c^ + c s e- 8x . 
 
 5. y = c l e~* x + c 2 . 9. y = c^ 1 + c z et x + c 8 e~ x . 
 
 10. y = ce* x + ce 2x + ce- 6a; . 
 
 EXERCISE 15 
 
 1. y = e 8:r (c 1 cosx + c 2 sinx). 3. y = c x cos2x + c 2 sin2x. 
 
 2. y = e* x (c t cos 2 x + c 2 sin 2 x). 4. ?/ = c t cos ox + c 2 sin ox. 
 
 a; / / / \ 
 
 5. y = Cje 2 * + e 2 (c 2 cos - x + c s sin - xj- 
 
 6. y = e ^(cjcos x + c 2 sin x). 
 
 \ 2i mi 
 
 7. y = c^ x + e~ x (c 2 cos V3 x + C 3 sin \/3 x) . 
 
 8. y = C 1 e 2x + c 2 e- 2a: + c 3 cos2x + c 4 sin2x. 
 
 + c 2 e 3 + ^(Cg cosx + c 4 sin x). 
 
 cos 
 2 
 
 x + c 2 sin x)+e~ 2 (c 3 cos x + c 4 sin x). 
 2 / \ 2 2/
 
 48 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 16 
 
 2. y = 0*(c l x + Cy). 
 
 8. y = e 2 *^ + c 2 ) + C 8 **. 
 
 4. y = e 2 *^ + c 2 ) + c 3 x + c 4 . V = ^('i* + C 2 ) + c ^ &x 
 
 5. r, = c 1 e5*+c 2 x 8 + c 3 x 2 + c 4 x + c s . ">. y = e 2 *^ 2 + c^ + c,). 
 
 6. y = e 8 * (qx 2 + CgX + c 8 ) + c 4 e**. 
 
 EXERCISE 17 
 
 1. y = ^(CjX* + c^ + c 3 ) + e 2a; . 
 
 2. y = CjC 8 * + c 2 cos \/3x + c 3 sin V3x e 2 *. 
 
 3. y = CT&* + c^^ + c 8 e^ + \ &*. 
 
 4. 7/ = ce 2a: + ce- 2a: + ccos2x + Csin2x 
 
 5. y = CjC 81 + c 2 e 9 * + 3x 5. 
 
 6. y = c 1 e* I + e^(c 2 -^x). 
 
 7. 2/ = c 1 e a: + 026-^+ 2x 2 3x 4. 
 
 8. y = Cj cos 2 x + c 2 sin 2 x + \ x cos 2 x. 
 
 150 442 
 
 2x x 
 
 10. y = c 1 e 8 +c 2 e 2 + c. 
 
 EXERCISE 18 
 
 1. xy = Cj logx + c 2 . 3. y = c a x + c 2 x 2 + c 8 x 8 . 
 
 X 2 C 
 
 2. y = c 1 x + c 2 x*- . 4. y = c 1 x 2 + c 2 x*+ -J. 
 
 5. y + x (c 2 cos V3 log x + c 3 sin V log x) + ^ logx. 
 
 6. x*y = c a x 8 + c^ 2 + CgX + C 4 . 8. ?/ = Cj(ax + 6)* + c 2 (ax + 6)~ 8 . 
 
 7. s = c 1 + c 2 -fclog. 9. 2/ = Ci(2x-3) + c 2 (2x-3)~i. 
 
 10. y = c a (x I) 2 + c 2 cos log (x I) 2 + c 8 sin log (x I) 2 . 
 
 EXERCISE 19 
 
 1. y = Cj logx + c 2 . 6. y = x + c 2 cos(x + c^. 
 
 2. y = Cj sin-ix + c 2 . 7. y ='c l sec~ 1 x + c 2 . 
 
 3. y ^= log sec (x + c t ) + c 2 . 8. y = c t sec x + c 2 . 
 
 4. i/ = log(x + Cj) + c 2 . 9. i/ = (x + c 1 )log(x + Cj) + c 2 . 
 
 5. y = c x tanx + c 2 . 10. (x - c a ) 2 + (y c ? ) 2 = a 2 .
 
 1. (X + Cl ) 2 + 2/ 2 = C 2 . 
 
 2. e^ = c x sin x + c 2 cosx. 
 
 3. y = CjeV. 
 
 4. log?/ = Cje 2 * + c 2 e- 2 *. 
 
 5. y8 = Cie 2x + C2e -2x. 
 
 ANSWERS 
 
 EXERCISE 20 
 
 49 
 
 6. y = c 1 sec(x + c 2 ). 
 
 7. 2/^ = C 1 sin(2x + c 2 )- 
 
 8. smy = CjC* + c 2 e- x . 
 
 9. i&ny = CjC" 5 + c^-*. 
 10. a sin- 1 ^ = x + c^ + c 2 . 
 
 EXERCISE 21 
 
 1* y uic/ T i^ 2 
 
 2. y = c t sin (ox + c 2 ). 
 
 3. (x + c 2 ) 2 - c 2 ?y 2 = c t 
 
 5. 
 
 6. Cj^ Vy _ 2 cf + 2 cj 5 log (y^ + Vy 2 cf) = x + c 2 . 
 
 ay 
 
 7. 4CjC 2 e 2 = 2c 1 2 e c 2 a: + cur* 1 **. 
 
 C 1 2 )=2x + c 2 . 
 
 9. 
 
 10. sin y = Vi + c 2 sin 
 
 EXERCISE 22 
 
 1. (x 2 3x)y = & + x s + c x x 2 + c 2 x + c 3 . 
 
 2. (2x + 5) 2 j/ = sinx + c^x + c 2 . 
 
 X 
 
 3. y sinx = 4 e 2 + qx + c 2 . 
 
 4. ?yVl + x 2 = (1 + x 2 )i + c t log(x + Vl + x 2 )+ C 2 . 
 
 6. (3x 2 - 5x + 7)i/ = e^ + CjX + c 2 . 
 
 7. x 7 y = CjX 2 + c^ + c 3 + 3 &*> + cosx. 
 
 8. y = cosx + e 2 * + c t x 2 + c^e + c g . 
 
 9. e^(y + CjX + c 2 ) = x + c 8 . 
 
 10. y = a sin- 1 + c t x + c 2 . 
 
 Cv
 
 50 
 
 DIFFERENTIAL EQUATIONS 
 
 EXERCISE 23 
 
 1. y = G! sin x sin (2 x + a). 
 
 ^ 
 
 2. y = e 2 (Cj cos3x + c 2 sinSz). 
 
 3. ?/ = sinx(c 1 e I + c 2 ). 
 
 4. y = e^c^x 4 + c 2 x 8 ). 
 
 5. y = x (c x cos 3 x + c 2 sin 3 x). 
 
 6. y sin ox = c 1 e ax + Cge-"*. 
 
 7. 7/ = e v ^(c 1 x + c 2 ). 
 
 8. y = xe* (Cj cos 2 x + c 2 sin 2 x). 
 
 9. y = logx (Cje* + c 2 e-*). 
 
 10. y log x = G! cos ox + c 2 sin ax. 
 
 EXERCISE 24 
 
 1. y = Cj sin [2 log (sec x + tan x) + a] . 
 
 2. y = c x x Vl - x 2 + c 2 (1 - 2 x 2 ). 
 
 3. y Vx 2 + 1 = CjX + c 2 . 
 
 4. y = Cj sec 2 x + c 2 cos 2 x. 
 
 5. xj/ = c a 4 c 2 Vx 2 1. 
 
 6. y 
 
 x 2 4 a 2 = c t x + c 2 . 
 
 7. ?/ = c x sin x + c 2 esc x. 
 
 aa- 2 aa* 
 
 8. y = c^e 2 + c.-.e~ 2 . 
 
 9. y = Cj sec x + c., tan x. 
 10. y = 
 
 EXERCISE 25 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 7. 
 
 8. 
 17. 
 18. 
 19. 
 20. 
 23. 
 24. 
 25. 
 
 w C 6"^ -4* C 6~ a " c . 
 
 y = cos(x4-c a ) + c 2 . 
 y = c 1 x + c 2 + <* x (c a x + c 4 ). 
 
 y = CjX + c 2 Vl x 2 . 
 y = Cj sec*x + c 2 cosx. 
 sin y = CjX + C 2 , 
 
 y = c x (logx) 2 + c 2 (logx) 8 . 
 y = e 1 (GJ cos x 4- c 2 sin x) 4 
 y= ( x (c l +x) + (* x (c 2 + x). 
 & c^fF + c 2 e- 
 
 (logy) 2 = c 1 e 2az 4- c 2 e~ Zax 
 
 10. y = e~ 2x (c 1 cos2x 4 c 2 sin2a-). 
 
 11. y = c 1 (2x-3) a 4-c 2 (2x-3)-. 
 
 12. y = & (c x sin x 4 o 2 ). 
 
 13. 2/ = c 1 4c 2 e6^4-c 3 e-3x. 
 
 14. y = c t tanx 4 c 2 . 
 
 15. y = GJ sin 2 x 4 c 2 sinx. 
 
 16. e^ = CjC" + c 2 e- . 
 
 c 4 sinx). 
 
 21. tan 1 y = c^e* 4- c 2 e~ I . 
 
 22. s = c t sin(oi + a). 
 C 2 sin3x)4- e 3l (c 3 cos2x 4 C 4 sin2x). 
 
 27. y = e 035 (c t cos ax + C 2 sin ax). 
 
 28. ^' = c 1 sinx 4 c 2 . 
 
 29. ?/ = Vx 2 4 a 2 4- CjX 4 c 2 .
 
 ANSWERS 
 
 51 
 
 EXERCISE 26 
 
 4. y-kx 2 +c. 
 
 2. ?/ 2 = 2 fee 3 + c. 5. y 2 = 
 
 3. y 2 = x 2 + c 2 . 6. y = ce kx . 
 9. (x-2/) 2 -2fc(x + ?/) + A; 2 = 0. 
 
 10. xy = A:' 2 . 11. xy c 2 . 
 
 13. qs 2 = k + (cj + c 2 ) 2 . 
 14. 
 
 15. s = c sin (fci + ) 
 
 16. (x - c,) 2 + (y - c 2 ) 2 - a 2 . 
 
 17. x + C 2 = c (2y-c. 
 
 18. y + c 2 = u 
 
 2 \ 8 
 
 7. 4x 2 + Qy* = c 2 . 
 
 8. x 2 +^ 2 = c 2 . 
 
 = fc 2 - (x + Cl ) 2 .] 
 
 384 
 
 P /Zx 2 x 8 \ Pi 8 
 20. Ely = I --- ) --- 
 2 \ 4 6/ 32 
 
 22. 
 
 24 
 
 p 
 
 -( 
 6 V 
 
 6 24 
 
 P/8 
 
 . 
 
 6 
 
 . C = - + ce 
 
 __ e JtK r _ 
 
 24. C = ce H+ - -- / eRKf'(C)dt. 
 
 K J 
 
 t 
 __L_ e ~~RK r _L_ 
 
 25. g = ce Ef+ " -- / e 
 
 It J
 
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