y = ± x 2 + c. ln y = 2 ln x + 2C sin(x) dx

Transcription

1 Worked Solutions Chapter 4: Separable First-Order Equations 43 a Factoring out 2, we get 3 sinx)) 2, which is f x)g), ds with f x) 3 sinx) and g) 2 So the equation is separable 43 c x x )2 x )2 f x)g) for an choice of f and g x So the differential equation is not separable 43 e 43 g 44 a ) f x)g) with f x) 4 and g) 2 So the differential equation is separable + 4 x 2 x 2 4 f x)g) for an choice of f and g ds So the differential equation is not separable x x x x x 2 + C 2 x 2 + }{{} 2C c 44 c ± x 2 + c [ x ][ ] x ) x x x 2 ln ln x + C ln ln x + 2C ±e 2 ln x +2C ±e 2 ln x e 2C ±e 2C e ln x2 Ax 2 44 e 2 Ax 2 9 ± Ax 2 9 cos) sinx) cos) sinx) sin) cosx) + c arcsinc cosx)) 45 a The general solution from the solution to exercise 44 a) is ± x 2 + c Appling the initial condition, we have 3 ) ± 2 + c ± + c Since 3 is positive, we must take the positive square root For c, we then have 3 + c c c 9 8 So the solution is x 2 + 8

2 2 Separable First-Order Equations 45 c Finding the general solution to the differential equation: x2 + x x 2 + ) x x x 2 ln 2 + ) 2 x 2 + C ln 2 + ) x 2 + c 2 + e x2 +c e x2 e c e x2 A 2 Ae x2 ± Ae x2 Appling the initial condition: 2 0) ± Ae 02 ± A So we take the negative square root, and then solve for A : 2 A 4 A A 5 So the solution is 5e x2 46 a 0 46 c x2 9x x 4x x 4) x2 9x x ± 9 ±3 So the two constant solutions are 3 and 3 46 e 0 ex+2 e x e 2 But there are no values of such that e 2 0 So there are no constant solutions 47 a From the answer to exercise 46 a, we know 4 is the onl constant solution To find the nonconstant solutions: x 4x x 4) 4 x 4 x ln 4 2 x 2 + c ) 4 2 x 2 + c e c 2 x 2) 4 ±e c 2 x 2) A 2 x 2) 4 + A 2 x 2) with A ±e c 0 ) Since, the last equation reduces to the constant solution 4 when A 0, that last equation without restrictions on A can serve as the general solution

3 Worked Solutions 3 47 c Constant solutions: 0 x 3x x 3) 2 3) x 2) 3) x 2) 3) x 2) 3) 3 x 2 3 [x 2] ln 3 2 x 2 2x + c ) ) 3 2 x 2 2x + c 2 x 2 2x e c ) 3 ±e c 2 x 2 2x ) 3 + Ae c 2 x 2 2x with A ±e c 0 Since, the last equation reduces to the constant solution 3 when A 0, that last equation without restrictions on A can serve as the general solution 47 e Constant solutions: 0 0 is the constant solution x Other solutions: x x x ln ln x + C ±e ln x +C ±e C e ln x Ax Since the last equation becomes the constant solution 0 when A 0, we can use that equation Ax, for the general solution 47 g x 2 + ) > 0 no constant solution x x x arctan) arctanx) + c x2 Solving for then ields the general solution tanarctanx) + c) 47 i There are no constant solutions since e > 0 for all So all solutions are nonconstant: e e e e x + c lnx + c)

4 4 Separable First-Order Equations 47 k 0 3x3 0 is the constant solution 3x3 3 3x x 2 + C 2 3x 2 + c ± c 3x 2) / 2 3x This last equation does not reduce to the constant solution 0 for an choice of c So, to describe the general solution we need both ± c 3x 2) / 2 and 0 47 m 3x 2 2 3x 2 3x2 3x 2 3x 2 2 ) Constant solutions: 0 3x 2 2 ) 2 0 ± and are the two constant solutions 3x 2 2 ) 2 3x 2 2 3x 2 x 3 + C ) For the remaining integral, ou can use partial fractions Begin b noting that 2 + ) ) A + + B A ) + B + ) + ) ) A ) + B + ) 2 So A and B are numbers such that A ) + B + ) for all Now solve for A and B, possibl b first setting, A ) + B + ) B 2, and then setting, A ) + B + ) A 2 So, 2 / / 2,

5 Worked Solutions 5 and ignoring the arbitrar constant) [ ] + 2 [ln ln + ] 2 ln + Combining this with equation ): 2 ln + x 3 + C + ±e2x3 +2C Ae 2x3 Ae 2x3 + Ae 2x3 Ae 2x3 + Ae 2x3 Ae 2x3) + Ae 2x3 + Ae2x3 Ae 2x3 If A 0, the last equation reduces to, but the equation does not reduce to for an choice of A So all the solutions are given b using both + Ae2x3 Ae 2x3 and 47 o ) Constant solutions: ) 0 and ) 00 ) 2 2x + c ) 00 ) Using partial fractions see the solution to exercise 47 m), 00 ) 00 [ ] + 00 [ln ln 00 ] ln 00 Combined with equation ), this gives 00 ln 00 2x + c 00 ±e002x+c) Ae 200x 00Ae 200x Ae 200x + Ae 200x) 00Ae 200x 00Ae200x + Ae 200x

6 6 Separable First-Order Equations The last reduces to 0 if A 0, but does not reduce to 00 for an choice of A So all the solutions are given b using both 00Ae200x + Ae 200x and a ) Clearl, the onl constant solution is 5 which does not satisf the initial condition 0) 8 So we must find the nonconstant solutions: 2 5) ln 5 2x + c 5 ±e 2s+c Ae 2x 5 + Ae 2x Appling the initial condition: 8 0) 5 + Ae A A So the solution to the initial-value problem is 5 + 3e 2x 48 c x 2 ) x Clearl, the onl constant solutions are 0 and, neither of which satisfies the initial condition ) 2 So we must find the nonconstant solutions: ) x ) x ) ln x + c ) x Using partial fractions see the solution to exercise 47 m), we find that So, continuing from equation ), we have ) ln ln ln ln x + c ln x + c ±e ln x +c Ax Ax Ax Ax Appling the initial condition: 2 ) A A 2 So, 2 2 x 2 x A 2

7 Worked Solutions 7 48 e 2 ) 4x 4x 2 The onl constant solution is 0, which does not satisf the initial condition 0) So we must find the nonconstant solutions: 2 ) 4x 2 4x [ 2 4x ] 4x 2 2 ln 2x 2 + c 2 2 ln 4x 2 + 2c Getting an licit solution here is not practical So we will appl the initial condition 0) to the last equation: 2 2 ln c c 2c 5 So the implicit) solution is 2 2 ln 4x a From the answer to exercise 48 a we know the solution is 5+3e 2x which is valid for all values of x So the interval is, ) 40 c From the answer to exercise 48 e, we know the solution is 2 2 x, which is continuous everwhere except at x 2 where it blows up And since the initial condition is given at x < 2, the interval over which this solution is valid is, 2) 40 e From the answer to exercise 47 k, we know all the solutions to the differential equation are given b ± c 3x 2) / 2 and 0 Obviousl, 0 does not satisf the initial condition So we appl the initial condition with the nonconstant solution formula: and 2 0) ± c 3 0 2) /2 ± c c 4, + 4 3x 2) / 2 4 3x 2 This is valid over the largest interval containing 0 and with 4 3x 2 > 0 But 4 3x 2 > 0 x 2 < < x < ) So the interval is 3 2, 3 2