CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS SECTION 1.1 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of differential equations, and to show the student what is meant by a solution of a differential equation Also, the use of differential equations in the mathematical modeling of real-world phenomena is outlined Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the given differential equations We include here just some typical examples of such verifications If y1 cos x and y2 sin x , then y1 2sin x y2 cos x , so y1 4 cos x 4 y1 and y2 4sin x 4 y2 Thus y1 y1 and y2 y2 If y1 e3 x and y2 e 3 x , then y1 e3 x and y2 e 3 x , so y1 9e3 x y1 and y2 9e 3 x y2 If y e x e x , then y e x e x , so y y e x e x e x e x e x Thus y y e x If y1 e 2 x and y2 x e 2 x , then y1 e 2 x , y1 e 2 x , y2 e 2 x x e 2 x , and y2 e 2 x x e 2 x Hence y1 y1 y1 e 2 x 2 e 2 x e 2 x and y2 y2 y2 4e 2 x x e 2 x e 2 x x e 2 x x e 2 x If y1 cos x cos x and y2 sin x cos x , then y1 sin x 2sin x, y1 cos x cos x, y2 cos x 2sin x , and y2 sin x cos x Hence y1 y1 cos x cos x cos x cos x 3cos x and y2 y2 sin x cos x sin x cos x 3cos x Copyright © 2015 Pearson Education, Inc 11 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS If y y1 x 2 , then y x 3 and y x 4 , so x y x y y x x 4 x 2 x 3 x 2 If y y2 x 2 ln x , then y x 3 x 3 ln x and y x 4 x 4 ln x , so x y x y y x 5 x 4 x 4 ln x x x 3 x 3 ln x x 2 ln x 5 x 2 x 2 x 2 10 x 2 x 2 ln x 13 Substitution of y erx into y y gives the equation 3r e rx e rx , which simplifies to r Thus r / 14 Substitution of y erx into y y gives the equation 4r e rx e rx , which simplifies to r Thus r / 15 Substitution of y erx into y y y gives the equation r e rx r e rx e rx , which simplifies to r r (r 2)(r 1) Thus r 2 or r 16 Substitution of y erx into y y y gives the equation 3r e rx 3r e rx e rx , which simplifies to 3r 3r The quadratic formula then gives the solutions r 3 57 The verifications of the suggested solutions in Problems 17-26 are similar to those in Problems 1-12 We illustrate the determination of the value of C only in some typical cases However, we illustrate typical solution curves for each of these problems 17 C2 18 C 3 Copyright © 2015 Pearson Education, Inc Section 1.1 Problem 17 Problem 18 (0, 3) (0, 2) y y −4 −4 0 −5 −5 x 19 If y x Ce x , then y gives C , so C 20 If y x C e x x , then y 10 gives C 10 , or C 11 Problem 19 Problem 20 10 20 y x (0, 5) (0, 10) y 0 −5 −10 −5 −20 −10 −5 x x 21 C 22 If y ( x) ln x C , then y gives ln C , so C Copyright © 2015 Pearson Education, Inc 10 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS Problem 21 Problem 22 10 (0, 7) y y 0 (0, 0) −5 −10 −2 −1 −5 −20 −10 x 23 If y ( x ) 14 x C x 2 , then y gives 14 32 C 18 , or C 56 24 C 17 Problem 23 30 20 20 10 10 y −10 −20 −20 (1, 17) −10 −30 −30 0.5 1.5 2.5 3.5 4.5 x 25 20 Problem 24 30 (2, 1) y 10 x x If y tan x C , then y gives the equation tan C Hence one value of C is C / , as is this value plus any integral multiple of Copyright © 2015 Pearson Education, Inc Section 1.1 Problem 25 Problem 26 10 y (0, 1) (, 0) y 0 −2 −4 −2 −5 −1 −10 x 26 10 x Substitution of x and y into y x C cos x yields C 1 , so C 27 y x y 28 The slope of the line through x, y and x 2, is y tial equation is xy y 29 y0 y x , so the differenxx/2 If m y is the slope of the tangent line and m is the slope of the normal line at ( x, y ), then the relation m m yields m 1 y y 1 x Solving for y then gives the differential equation 1 y y x 30 31 Here m y and m Dx ( x k ) x , so the orthogonality relation m m gives the differential equation xy The slope of the line through x, y and ( y , x ) is y x y y x , so the differential equation is ( x y ) y y x In Problems 32-36 we get the desired differential equation when we replace the “time rate of change” of the dependent variable with its derivative with respect to time t, the word “is” with the = sign, the phrase “proportional to” with k, and finally translate the remainder of the given sentence into symbols 32 dP dt k P 33 dv dt kv Copyright © 2015 Pearson Education, Inc DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS 34 dv dt k 250 v 36 dN dt kN P N 35 dN dt k P N 37 The second derivative of any linear function is zero, so we spot the two solutions y x and y ( x) x of the differential equation y 38 A function whose derivative equals itself, and is hence a solution of the differential equation y y , is y ( x ) e x 39 We reason that if y kx , then each term in the differential equation is a multiple of x The choice k balances the equation and provides the solution y ( x) x 40 If y is a constant, then y , so the differential equation reduces to y This gives the two constant-valued solutions y ( x ) and y ( x) 1 41 We reason that if y ke x , then each term in the differential equation is a multiple of e x The choice k 12 balances the equation and provides the solution y ( x ) 12 e x 42 Two functions, each equaling the negative of its own second derivative, are the two solutions y x cos x and y ( x ) sin x of the differential equation y y 43 (a) We need only substitute x(t ) C kt in both sides of the differential equation x kx for a routine verification (b) The zero-valued function x (t ) obviously satisfies the initial value problem x kx , x(0) 44 (a) The figure shows typical graphs of solutions of the differential equation x 12 x (b) The figure shows typical graphs of solutions of the differential equation x 12 x We see that—whereas the graphs with k 12 appear to “diverge to infinity”—each solution with k 12 appears to approach as t Indeed, we see from the Problem 43(a) solution x(t ) C 12 t that x(t ) as t 2C However, with k 12 it is clear from the resulting solution x(t ) C 12 t that x(t ) remains bounded on any bounded interval, but x(t ) as t Copyright © 2015 Pearson Education, Inc Section 1.1 Problem 44a Problem 44b 4 x x 2 1 t 45 t , so Substitution of P and P 10 into the differential equation P kP gives k 100 t The initial condition Problem 43(a) yields a solution of the form P(t ) C 100 P (0) now yields C 12 , so we get the solution P (t ) 100 50 t t 100 We now find readily that P 100 when t 49 and that P 1000 when t 49.9 It appears that P grows without bound (and thus “explodes”) as t approaches 50 46 Substitution of v 1 and v into the differential equation v kv gives k 251 , so Problem 43(a) yields a solution of the form v(t ) C t 25 The initial condition v(0) 10 now yields C 101 , so we get the solution v(t ) 50 2t t 10 25 We now find readily that v when t 22.5 and that v 0.1 when t 247.5 It appears that v approaches as t increases without bound Thus the boat gradually slows, but never comes to a “full stop” in a finite period of time 47 (a) y (10) 10 yields 10 C 10 , so C 101 10 (b) There is no such value of C, but the constant function y ( x) satisfies the conditions y y and y (0) Copyright © 2015 Pearson Education, Inc DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS (c) It is obvious visually (in Fig 1.1.8 of the text) that one and only one solution curve passes through each point (a, b) of the xy-plane, so it follows that there exists a unique solution to the initial value problem y y , y (a ) b 48 (b) Obviously the functions u ( x) x and v ( x) x both satisfy the differential equation xy y But their derivatives u ( x) 4 x and v( x) 4 x match at x , where both are zero Hence the given piecewise-defined function y x is differentiable, and therefore satisfies the differential equation because u x and v x so (for x and x , respectively) (c) If a (for instance), then choose C fixed so that C a b Then the function C x if x y x C x if x satisfies the given differential equation for every real number value of C SECTION 1.2 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS This section introduces general solutions and particular solutions in the very simplest situation — a differential equation of the form y f x — where only direct integration and evaluation of the constant of integration are involved Students should review carefully the elementary concepts of velocity and acceleration, as well as the fps and mks unit systems Integration of y x yields y ( x) x 1 dx x x C Then substitution of x , y gives C C , so y x x x Integration of y x yields y x x dx 13 x C Then substitution 2 of x , y gives C C , so y x 13 x 3 Integration of y x yields y x y gives 163 C , so y x x 3/2 x dx 23 x 3/2 C Then substitution of x , 8 Integration of y x 2 yields y x x 2 dx 1 x C Then substitution of x , y gives 1 C , so y x 1 x Copyright © 2015 Pearson Education, Inc Section 1.2 Integration of y x 1 yields y x x 1 dx x C Then substitu- tion of x , y 1 gives 1 C , so y x x Integration of y x x 12 yields y x x x dx 13 x 12 substitution of x 4 , y gives 13 (5)3 C , so y x 13 x 32 3/2 C Then 125 10 10 yields y x dx 10 tan 1 x C Then substitution of x 1 x 1 x , y gives 10 C , so y x 10 tan 1 x Integration of y Integration of y cos x yields y x cos x dx 12 sin x C Then substitution of x , y gives C , so y x 12 sin x yields y ( x) Integration of y 10 Integration of y xe x yields 1 x 1 x 1 x , y gives C , so y x sin x 2 dx sin 1 x C Then substitution of y x xe x dx ueu du u 1 eu x 1 e x C , using the substitution u x together with Formula #46 inside the back cover of the textbook Then substituting x , y gives C , so y ( x) ( x 1) e x 11 If a t 50 , then v t 50dt 50t v0 50t 10 Hence x t 50t 10 dt 25t 10t x0 25t 10t 20 12 If a t 20 , then v t 20 dt 20t v0 20t 15 Hence x t 20t 15 dt 10t 15t x0 10t 15t 13 If a t 3t , then v t 3t dt 32 t v0 32 t Hence x t 14 t dt 12 t 5t x0 12 t 5t 2 If a t 2t , then v t 2t 1 dt t t v0 t t Hence x t t t dt 13 t 12 t 7t x0 13 t 12 t 7t Copyright © 2015 Pearson Education, Inc 10 15 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS If a t t 3 , then v t t 3 dt 2 t 3 C t 3 3 37 (taking C 37 so that v 1 ) Hence x t 16 t 3 37 dt 13 t 3 37t C 13 t 3 37t 26 , then v t t4 that v 1 ) Hence If a t dt t C t (taking C 5 so t4 x t t dt t 4 3/2 5t C t 4 3/2 5t 293 (taking C 29 so that x ) 17 If a t t 1 , then v t t 1 dt 12 t 1 C 12 t 1 12 (taking 3 3 C 2 2 so that v ) Hence x t 12 t 1 12 dt 2 t 1 1 1 12 t C 12 t 1 t 1 (taking C 12 so that x ) 18 If a t 50sin 5t , then v t 50 sin 5t dt 10 cos 5t C 10 cos 5t (taking C so that v 10 ) Hence x t 10 cos 5t dt 2 sin 5t C 2 sin 5t 10 (taking C 10 so that x ) Students should understand that Problems 19-22, though different at first glance, are solved in the same way as the preceding ones, that is, by means of the fundamental theorem of calculus in the form x t x t0 v s ds cited in the text Actually in these problems x t v s ds t t t0 , since t0 and x t0 are each given to be zero 19 if t 5 The graph of v t shows that v t , so that 10 t if t 10 if t 5t C1 x t Now C1 because x , and continuity of 10t t C2 if t 10 x t requires that x t 5t and x t 10t 12 t C2 agree when t This implies that C2 252 , leading to the graph of x t shown Copyright © 2015 Pearson Education, Inc Section 1.6 85 y y2 ln ln x ln 2001 10 x x 10 Exponentiation and then multiplication of the resulting equation by x finally leads to y x y 200 x 10 , as desired Note that if x , then this equation yields y , confirming that the airplane reaches the airport at the origin 71 Equations (12)-(19) apply to this situation as with the airplane in flight w (a) With a 100 and k , the solution given by Equation (19) is v0 x 1 x 3 y 50 The fact that y 0 means that this trajectory goes 100 100 through the origin where the tree is located (b) With k x , the solution is y 50 1 , and we see that the dog hits the 100 bank at a distance y 0 50ft north of the tree (c) With k x 1 x , the solution is y 50 This trajectory is as4 100 100 ymptotic to the positive x-axis, so we see that the dog never reaches the west bank of the river 72 We note that the dependent variable y is missing in the given differential equation ry 1 y 32 , leading us to substitute y , and y This results in rp 1 , a separable first-order differential equation for as a function of x 32 Separating variables gives r 1 32 d dx , and then integral formula #52 in the back of our favorite calculus textbook gives r 1 2 r 1 x a We solve readily for dy 2 dx xa r x a y x a , that is, x a , so that r2 x a Finally, a further integration gives x a dx r2 x a r2 x a b , Copyright © 2015 Pearson Education, Inc 86 SUBSTITUTION METHODS AND EXACT EQUATIONS which leads to x a y b r , as desired 2 CHAPTER Review Problems The main objective of this set of review problems is practice in the identification of the different types of first-order differential equations discussed in this chapter In each of Problems 1–36 we identify the type of the given equation and indicate one or more appropriate method(s) of solution y x , showing that the x equation is linear An integrating factor is given by exp dx e 3ln x x 3 , and x 3 4 1 multiplying the equation by gives x y 3x y x , or Dx x 3 y x 1 Integrat- We first rewrite the differential equation for x as y ing then leads to x 3 y ln x C , and thus to the general solution y x ln x C y x , showing that the y2 x equation is separable Separating variables yields ln x C , and thus the geny x x 1 eral solution y 3 ln x C x ln x C x We first rewrite the differential equation for x, y as xy y y y shows that the x2 x x equation is homogeneous Actually the equation is identical to Problem in Section 1.6; x the general solution found there is y C ln x Rewriting the differential equation for x as y Rewriting the differential equation in differential form gives M dx N dy xy e x dx x y sin y dy , M x, y xy N x, y , the given equation is exact Thus we apy x ply the method of Example in Section 1.6 to find a solution of the form F x, y C and because First, the condition Fx M implies that F x, y xy e x dx x y e x g y , Copyright © 2015 Pearson Education, Inc Review Problems 87 and then the condition Fy N implies that 3x y g y 3x y sin y , or g y sin y , or g y cos y Thus the solution is given by x y e x cos y C y x , showing that the y x4 2x dx , or equation is separable Separating variables yields dy y x4 1 1 x 1 x ln y C C , leading to the general solution y C exp , x x x x where C is an arbitrary nonzero constant We first rewrite the differential equation for x, y as We first rewrite the differential equation for x, y as equation is separable Separating variables yields y x , showing that the y2 x2 1 2x dy dx , or y x2 1 1 x ln x Cx ln x C , that is , leading to the general solution y x y x x y x ln x Cx y , showing that the x x equation is linear An integrating factor is given by exp dx e ln x x , and x 1 multiplying the equation by gives x y xy , or Dx x y Integrating x x ln x C then leads to x y ln x C , and thus to the general solution y x2 We first rewrite the differential equation for x as y dy y y , showing that is it dx x x y dv dv homogeneous Substituting v then gives v x v 2v , or x v 3v Sepx dx dx 1 dv dx Upon partial fraction decomposition the arating variables leads to v 3v x 1 v3 solution takes the form ln v ln v ln x C , or C x , where C is an 3 v arbitrary positive constant, or v Cvx , where C is an arbitrary nonzero constant We first rewrite the differential equation for x as Copyright © 2015 Pearson Education, Inc 88 REVIEW PROBLEMS Back-substituting finally y y y y for v then gives the solution C x , or y 3x Cyx , or x x x 3x Cx dy y y shows that it is a Bernoulli dx x x 1 1 equation with n The substitution v y y implies that y v 1 and thus that 2 y v 2 v Substituting gives v 2v v 1 v 2 , or v v , a linear equax x x x tion for v as a function of x An integrating factor is given by exp dx x 2 , x and multiplying the differential equation by gives v v , or x x x 1 1 Cx Finally, Dx v Integrating then leads to v C , or v 3x x 3x x x 1 back-substituting y 1 for v gives the general solution Cx , or y 3x 3x , as found above y Cx Cx 3x Alternatively, writing the given equation as y xy1 , showing that it x 12 is a Bernoulli equation with n The substitution v y implies that y v and thus that y 2vv Substituting gives 2vv v xv , or v v 3x , a linear equax x tion for v as a function of x An integrating factor is given by exp dx x , and x multiplying the differential equation by gives xv v x , or Dx xv 3x InteWe first rewrite the differential equation for x, y as y C Finally, back-substituting y1 for v x C C x , or y x x x grating then leads to xv x C , or v x gives the general solution y1 Copyright © 2015 Pearson Education, Inc Review Problems 10 dy 1 x 1 y , showing that the equation is dx 1 separable Separating variables gives dy x dx , or tan 1 y x x C , 1 y or finally y tan x x C Factoring the right-hand side gives 11 12 89 dy y y , showing that it dx x x y dv dv v 3v , or x 3v Sepais homogeneous Substituting v then gives v x x dx dx 1 Backrating variables leads to dv dx , or 3ln x C , or v C 3ln x v x v x y y , or y for v then gives the solution substituting C 3ln x x x C 3ln x Alternatively, writing the equation in the form y y y for x, y shows that it x x is also a Bernoulli equation with n The substitution v y 1 implies that y v 1 and 3 thus that y v 2 v Substituting gives v 2v v 1 v 2 , or v v , a linx x x x ear equation for v as a function of x An integrating factor is given by exp dx x , and multiplying the differential equation by gives xv v x x 3ln x C Final, or Dx xv Integrating then leads to xv 3ln x C , or v x x ly, back-substituting y 1 for v gives the same general solution as found above We first rewrite the differential equation for x, y as Rewriting the differential equation in differential form gives xy y dx x y xy dy , xy y 18 xy y x y xy , the given equation is y x exact We apply the method of Example in Section 1.6 to find a solution in the form F x, y C First, the condition Fx M implies that and because F x, y xy y dx 3x y xy g y , and then the condition Fy N implies that x y xy g y x y xy , or g y , that is, g y is constant Thus the solution is given by 3x y xy C Copyright © 2015 Pearson Education, Inc 90 13 REVIEW PROBLEMS y x x , showing that the y equation is separable Separating variables yields dy x x dx , or y 1 x x C , leading to the general solution y C x x5 y We first rewrite the differential equation for y as 14 dy y y We first rewrite the differential equation for x, y as , showing that it is dx x x y dv dv v v , or x v Separathomogeneous Substituting v then gives v x x dx dx 1 1 Backing variables leads to dv dx , or ln x C , or v v x 2v C ln x y x2 y , or y for v then gives the solution substituting x C ln x x C ln x 1 y y for x, y shows that x x it is also a Bernoulli equation with n The substitution v y 2 implies that y v 1 1 1 and thus that y v 3 2v Substituting gives v 3 2v v 1 v 3 , or x x 2 v v , a linear equation for v as a function of x An integrating factor is given by x x exp dx x , and multiplying the differential equation by gives x 2 x v x v , or Dx x v Integrating then leads to x v ln x C , or x x ln x C v Finally, back-substituting y 2 for v gives the same general solution as x2 found above Alternatively, writing the equation in the form y 15 This is a linear differential equation An integrating factor is given by exp dx e3 x , and multiplying the equation by gives e3 x y 3e3 x y 3x , or Dx e3 x y x Integrating then leads to e3 x y x C , and thus to the general solu- tion y x C e 3 x 16 Rewriting the differential equation as y x y suggests the substitution v x y , which implies that y x v , and thus that y v Substituting gives v v , or v v , a separable equation for v as a function of x Separating variables gives Copyright © 2015 Pearson Education, Inc Review Problems 1 v dv dx , or (via the method of partial fractions) 91 ln v ln v x C , 1 v x C , or v Ce2 x 1 v Finally, back-substituting x y for v gives 1 v the implicit solution x y Ce2 x 1 x y or ln 17 Rewriting the differential equation in differential form gives e x ye xy dx e y xe xy dy , x e ye xy xye xy e y xe xy , the given equation is exact We ap y x ply the method of Example in Section 1.6 to find a solution in the form F x, y C and because First, the condition Fx M implies that F x, y e x ye xy dx e x e xy g y , and then the condition Fy N implies that xe xy g y e y xe xy , or g y e y , or g y e y Thus the solution is given by e x e xy e y C 18 dy y y , showing that it We first rewrite the differential equation for x, y as dx x x y dv dv 2v v , or x v v is homogeneous Substituting v then gives v x x dx dx 1 dv dx , or (after decomposing into partial fracSeparating variables leads to vv x 1 v2 dv dx , or ln tions) ln x C , or v 1 v 1 v x 1 v 1 v v2 1 v 1 v Cx , or v Cx 1 v 1 v Back-substituting y for v then gives the x y y y solution Cx , or finally y Cx x y x y Cx x y x x x Alternatively, rewriting the differential equation for x as y y y shows x x 1 2 that it is Bernoulli with n The substitution v y y implies that y v 1 , and 1 thus that y v 3 2v Substituting gives v 3 2v v 1 v 3 , or x x v v , a linear equation for v as a function of x An integrating factor is given by x x Copyright © 2015 Pearson Education, Inc 92 REVIEW PROBLEMS dx x , and multiplying the differential equation by gives x x v x 3v x , or Dx x v x Integrating then leads to x v x C , or exp x C Finally, back-substituting y 2 for v gives the general solution x x2 1 C 4 x y , or (for C ) C x x C , or 2 x y y x x x y y x x y Cx x y , the same general solution found above (Note that C the case C in this latter solution corresponds to the solutions y x , which are singular for the first solution method, since they cause v v to equal zero.) v 19 We first rewrite the differential equation for x, y as the equation is separable Separating variables yields y x 3 3x , showing that y dy x 3 3x dx , or y2 1 x2 2 x x C , leading to the general solution y 2 x x C x Cx y 20 y 3x 5 , showing that the x equation is linear An integrating factor is given by exp dx e3ln x x , and x 12 multiplying the equation by gives x y 3x y 3x , or Dx x y x1 InteWe first rewrite the differential equation for x as y grating then leads to x y x C , and thus to the general solution y x 3 Cx 3 21 1 y , showing that x 1 x 1 the equation is linear An integrating factor is given by exp dx x , and x 1 1 multiplying the equation by gives x 1 y y In, or Dx x 1 y x 1 x 1 tegrating then leads to x 1 y ln x 1 C , and thus to the general solution We first rewrite the differential equation for x as y y ln x 1 C x 1 Copyright © 2015 Pearson Education, Inc Review Problems 22 93 dy y 12 x y shows that it is a Bernoulli dx x equation with n The substitution v y12 y1 implies that y v and thus that y 3v v Substituting gives 3v 2v v 12 x 3v , or v v x , a linear equation x x for v as a function of x An integrating factor is given by exp dx x 2 , and x 2 3 multiplying the differential equation by gives x v x v x , or Dx x 2 v x Writing the given equation for x as Integrating then leads to x 2 v x C , or v x Cx Finally, back-substituting y for v gives the general solution y1 x Cx , or y x Cx 23 Rewriting the differential equation in differential form gives e y y cos x dx xe y sin x dy , y e y cos x e y cos x xe y sin x , the given equation is exact y x We apply the method of Example in Section 1.6 to find a solution in the form F x, y C First, the condition Fx M implies that and because F x, y e y y cos x dx xe y y sin x g y , and then the condition Fy N implies that xe y sin x g y xe y sin x , or g y , that is, g is constant Thus the solution is given by xe y y sin x C 24 We first rewrite the differential equation for x, y as the equation is separable Separating variables yields y x 3 x1 , showing that y dy x 3 x1 dx , or y x1 32 1 2 x x C , leading to the general solution y y x Cx1 25 y , showing that the x 1 2 equation is linear An integrating factor is given by exp dx x 1 , and x 1 2 multiplying the equation by gives x 1 y x 1 y x 1 , or We first rewrite the differential equation for x 1 as y Copyright © 2015 Pearson Education, Inc 94 REVIEW PROBLEMS 2 Dx x 1 y x 1 Integrating then leads to x 1 y x 1 C , and thus C to the general solution y x x 1 26 Rewriting the differential equation in differential form gives 9 x 12 y 12 x1 y dx x 3/2 y1/3 15 x 6/5 y1/2 dy , and because x1 y 12 x1 y 12 x1 y 18 x1 y1 x 3/2 y1/3 15 x 6/5 y1/2 , y x the given equation is exact We apply the method of Example in Section 1.6 to find a solution in the form F x, y C First, the condition Fx M implies that F x, y x1 y 12 x1 y dx x y 10 x y g y , and then the condition Fy N implies that x y1 15 x y1 g y x 3/2 y1/3 15x 6/5 y1/2 , or g y , that is, g is constant Thus the solution is given by x y 10 x y C 27 dy x2 y y shows that it is a Bernoulli dx x 3 equation with n The substitution v y implies that y v 1 and thus that Writing the given equation for x as 1 x2 y v 4 3v Substituting gives v 4 3v v 1 v 4 , or v v x , a line3 x 3 x ar equation for v as a function of x An integrating factor is given by exp dx x 3 , and multiplying the differential equation by gives x 3 4 x v x v x 1 , or Dx x 3 v x 1 Integrating then leads to x 3 v ln x C , or v x ln x C Finally, back-substituting y 3 for v gives the general solution y x 1 ln x C 28 1 2e x y , showing that the x x equation is linear An integrating factor is given by exp dx x , and multiply x We first rewrite the differential equation for x as y Copyright © 2015 Pearson Education, Inc Review Problems 95 ing the equation by gives x y y 2e2 x , or Dx x y 2e2 x Integrating then leads to x y e x C , and thus to the general solution y x 1 e x C 29 1 12 y x 1 , showas y 2x ing that the equation is linear An integrating factor is given by 12 dx x 1 , and multiplying the equation by gives exp 2x 12 1 12 x 1 y x 1 y x , or Dx x 1 y x Integrating then 12 leads to x 1 y x x C , and thus to the general solution We first rewrite the differential equation for x y x x C x 1 30 1 The expression x y suggests the substitution v x y , which implies that y v x , and thus that y v Substituting gives v v , or v v , a separable equa1 dv dx The further tion for v as a function of x Separating variables gives v 1 substitution v u (so that dv 2u du ) and long division give 2u dv du du 2u ln u 1 v ln u 1 u 1 v 1 leading to v ln v 1 , v x C Finally, back-substituting x y for v leads to the implicit general solution x x y ln 31 x y 1 C Rewriting the differential equation as y 3x y 21x shows that it is linear An integrating factor is given by exp gives e x y 3x 2e x y 21x 2e x 3 3x dx e , and multiplying the equation by , or D e y 21x e Integrating then leads to x3 x3 x3 x e x y 7e x C , and thus to the general solution y 7 Ce x 3 Alternatively, writing the equation for y 7 as dy 3x dx shows that it is separay7 ble Integrating yields the general solution ln y x C , that is, y Ce x , as found above (Note that the restriction y 7 in the second solution causes no loss of generality The general solution as found by the first method shows that either y 7 for all x or y 7 for all x Of course, the second solution could be carried out under the assumption y 7 as well.) Copyright © 2015 Pearson Education, Inc 96 32 REVIEW PROBLEMS dy x y y , showing that the equation is dx dy x dx , and the method of separable For y separating variables gives y y partial fractions yields We first rewrite the differential equation as 1 1 dy dy ln y y y y 1 y 1 y2 1 x C , or y Cy 2e x , or finally y leading to the solution ln y Ce x y2 , y dy xy xy shows that it is a Bernoulli dx 2 equation with n The substitution v y for y implies that y v 1 and thus 1 that y v 3 2v Substituting gives v 3 2v xv 1 xv 3 , or v xv 2 x , a 2 linear equation for v as a function of x An integrating factor is given by exp x dx e x , and multiplying the differential equation by gives Alternatively, writing the given equation as e x v xe x v 2 xe x , or Dx e x v 2 xe x Integrating then leads to 2 2 e x v e x C , or v Ce x Finally, back-substituting y 2 for v gives the same general solution as found above 33 2 Rewriting the differential equation for x, y in differential form gives 3x y dx xy dy , 3x y y xy , the given equation is exact We apply the y x method of Example in Section 1.6 to find a solution in the form F x, y C First, the and because condition Fx M implies that F x, y 3x y dx x xy g y , and then the condition Fy N implies that xy g y xy , or g y , that is, g is constant Thus the solution is given by x xy C dy 3x 1y shows that it dx 4y 2x y dv v , or is homogeneous Substituting v then gives v x x dx 4v Alternatively, rewriting the given equation for x, y as Copyright © 2015 Pearson Education, Inc Review Problems dv 3 6v v Separating variables leads to dx 4v 4v 97 4v dv dx , or 3 x y for v then gives the ln 6v 3 3ln x C , or 2v 1 x C Back-substituting x y 2 solution 1 x C , or finally y x x C , as found above x Still another solution arises from writing the differential equation for x, y as dy 3x y y 1 , which shows that it is Bernoulli with n 1 The substitution dx x v y implies that y v1 and thus that y v 1 2v Substituting gives 1 3x 3x v v v1 v 1 , or v v , a linear equation for v as a function of x 2x x An integrating factor is given by exp dx x , and multiplying the differential x 3x 3x equation by gives x v v , or Dx x v Integrating then leads to 2 x3 x v C Finally, back-substituting y for v leads to the general solution x3 x y C , that is, 2xy x C , as found above x 34 6v Rewriting the differential equation in differential form gives x y dx 3x y dy , x y 3x y , the given equation is exact We apply the y x method of Example in Section 1.6 to find a solution in the form F x, y C First, the and because condition Fx M implies that F x, y x y dx x 3xy g y , and then the condition Fy N implies that 3x g y 3x y , or g y y , or 1 g y y Thus the solution is given by x 3xy y C , or 2 2 x xy y C Copyright © 2015 Pearson Education, Inc 98 REVIEW PROBLEMS y 1 dy x shows that it is Alternatively, rewriting the given equation for x, y as y dx 3 x y dv 3v dv v 6v homogeneous Substituting v then gives v x , or x x dx v3 dx v v3 dv dx , or ln v 6v 1 2 ln x C Separating variables leads to v 6v x y for v then gives the solution , or x v 6v 1 C Back-substituting x y xy x C found above 35 dy 2x y 1 shows that it is separable For dx x 1 2x dy dx , or ln y 1 ln x 1 C , y 1 x 1 Rewriting the differential equation as y 1 separating variables gives leading to the general solution y C x 1 2x 2x y shows that it is x 1 x 1 2x , and multiplylinear An integrating factor is given by exp dx x 1 x 1 2x 2x , or ing the equation by gives y y 2 2 x 1 1 x x Alternatively, writing the differential equation as y 1 2x y C , or thus to the Integrating then leads to Dx y x 1 x 1 x x 1 general solution y 1 C x 1 found above 36 Rewriting the differential equation for x , y as dy cot x dx shows yy that it is separable The substitution y u gives dy du ln 1 u ln y , 1 u yy leading to the general solution ln y ln sin x C , or sin x y C , or finally y C csc x 1 Copyright © 2015 Pearson Education, Inc Review Problems 99 Alternatively, writing the differential equation for x , y as dy cot x y cot x y shows that it is Bernoulli with n The substitution dx 12 v y implies that y v and thus that y 2v v Substituting gives 1 2v v cot x v cot x v , or v cot x v cot x , a linear equation for v as a 2 1 function of x An integrating factor is given by exp cot x dx sin x , and mul2 tiplying the differential equation by gives sin x v sin x cot x v sin x cot x , sin x cot x Integrating then leads to sin x v sin x C , or or Dx sin x v v C csc x Finally, back-substituting y1 for v leads to the general solution y C csc x found above Copyright © 2015 Pearson Education, Inc