Separation of variables now shows that e y dy c 4 e y dt and then 1 2 c 4 e y 1

Separation of variables now shows that e y dy c 4 e y

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Separation of variables now shows that ± e y dy/ c + 4 e y = dt and then ± 1 2 ( c + 4 e y ) 1 / 2 = t + d . Algebraic manipulations then yield the implicitly defined solution e y = ( t + c 2 ) 2 + c 1 . 48. Suppose that y 0 = v ( y ) and then y 00 = v 0 ( y ) v ( y ). The equation is v 2 v 0 = 2,
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90 Chapter 2. First Order Differential Equations which gives us v 3 / 3 = 2 y + c . Now plugging 0 in place of t gives that 2 3 / 3 = 2 · 1 + c and we get that c = 2 / 3. This turns into v 3 = 6 y + 2, i.e. y 0 = (6 y + 2) 1 / 3 . This separable equation gives us (6 y + 2) - 1 / 3 dy = dt , and integration shows that 1 6 3 2 (6 y + 2) 2 / 3 = t + d . Again, plugging in t = 0 gives us d = 1 and the solution is (6 y + 2) 2 / 3 = 4( t + 1). Solving for y here yields y = 4 3 ( t + 1) 3 / 2 - 1 3 . 49. Set y 0 = v ( y ). Then y 00 = v 0 ( y )( dy/dt ) = v 0 ( y ) v ( y ). We obtain the equation v 0 v - 3 y 2 = 0, where the differentiation is with respect to y . Separation of variables gives vdv = 3 y 2 dy , and after integration this turns into v 2 / 2 = y 3 + c . The initial conditions imply that c = 0 here, so ( y 0 ) 2 = v 2 = 2 y 3 . This implies that y 0 = 2 y 3 / 2 (the sign is determined by the initial conditions again), and this separable equation now turns into y - 3 / 2 dy = 2 dt . Integration yields - 2 y - 1 / 2 = 2 t + d , and the initial conditions at this point give that d = - 2. Algebraic manipulations find that y = 2(1 - t ) - 2 . 50. Set v = y 0 , then v 0 = y 00 . The equation with this substitution is (1 + t 2 ) v 0 + 2 tv = ((1 + t 2 ) v ) 0 = - 3 t - 2 . Integrating this we get that (1 + t 2 ) v = 3 t - 1 + c , and c = - 5 from the initial con- ditions. This means that y 0 = v = 3 / ( t (1 + t 2 )) - 5 / (1 + t 2 ) . The partial fraction decomposition of the first expression shows that y 0 = 3 /t - 3 t/ (1 + t 2 ) - 5 / (1 + t 2 ) and then another integration here gives us that y = 3 ln t - 3 2 ln(1 + t 2 ) - 5 arctan t + d . The initial conditions identify d = 2 + 3 2 ln 2 + 5 π/ 4, and we obtained the solution. 51. Set v = y 0 , then v 0 = y 00 . The equation with this substitution is vv 0 = t . Inte- grating this separable differential equation we get that v 2 / 2 = t 2 / 2 + c , and c = 0 from the initial conditions. This implies that y 0 = v = t , so y = t 2 / 2 + d , and the initial conditions again imply that the solution is y = t 2 / 2 + 3 / 2.
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