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Unformatted text preview: Chapter 2 12. We have 3 vdv 1 4 v 2 = dx x Z 3 vdv 1 4 v 2 = Z dx x  3 8 Z du u = Z dx x ( u = 1 4 v 2 , du = 8 vdv )  3 8 ln 1 4 v 2 = ln  x  + C 1 1 4 v 2 = exp 8 3 ln  x  + C 1 = Cx 8 / 3 , where C = e C 1 is any nonzero constant. Separating variables, we lost constant solutions satisfying 1 4 v 2 = 0 v = 1 2 , which can be included in the above formula by letting C = 0. Thus, v = 1 Cx 8 / 3 2 , C arbitrary , is a general solution to the given equation. 14. Separating variables, we get dy 1 + y 2 = 3 x 2 dx Z dy 1 + y 2 = Z 3 x 2 dx arctan y = x 3 + C y = tan ( x 3 + C ) , where C is any constant. Since 1 + y 2 6 = 0, we did not lose any solution. 16. We rewrite the equation in the form x (1 + y 2 ) dx + e x 2 ydy = 0 , separate variables, and integrate. e x 2 xdx = ydy 1 + y 2 Z e x 2 xdx = Z ydy 1 + y 2 Z e u du = dv v ( u = x 2 , v = 1 + y 2 )  e u = ln  v  + C ln ( 1 + y 2 ) e x 2 = C is an implicit solution to the given equation. Solving for y yields y = p C 1 exp [exp ( x 2 )] 1 , where C 1 = e C is any positive constant, 36 Exercises 2.2 18. Separating variables yields dy 1 + y 2 = tan xdx Z dy 1 + y 2 = Z tan xdx arctan y = ln  cos x  + C. Since y (0) = 3, we have arctan 3 = ln cos 0 + C = C C = 3 ....
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 Spring '08
 MAZMANI
 Equations

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