Solutions manual for Advanced engineering mathematics 8ed by Erwin Kreyszig

By Erwin Kreyszig

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We find ∂ ∂M = [y cos(xy)] = cos(xy) − xy sin(xy), ∂y ∂y ∂ ∂N = [x cos(xy)] = cos(xy) − xy sin(xy). ∂x ∂x Therefore, the equation is exact and F (x, y) = x cos(xy) − y −1/3 dy = sin(xy) − 3 2/3 y + h(x) 2 ∂F 2 = y cos(xy) + h (x) = √ + y cos(xy) ∂x 1 − x2 2 ⇒ h (x) = √ ⇒ h(x) = 2 arcsin x, 1 − x2 and a general solution is given by sin(xy) − 3 2/3 y + 2 arcsin x = C. 2 22. In Problem 16, we found that a general solution to this equation is exy − xy −1 = C. Substituting the initial condition, y(1) = 1, yields e − 1 = C.

Linear. 6. Linear. 8. Writing the equation in standard form, dy y − = 2x + 1, dx x we see that P (x) = − 1 x ⇒ µ(x) = exp − 1 x dx = exp (− ln x) = 1 . x 2+ 1 x dx = x (2x + ln |x| + C) . Multiplying the given equation by µ(x), we get d y 1 =2+ dx x x ⇒ y=x 41 Chapter 2 10. From the standard form of the given equation, dy 2 + y = x−4 , dx x we find that (2/x)dx = exp (2 ln x) = x2 µ(x) = exp ⇒ d x2 y = x−2 dx y = x−2 ⇒ x−2 dx = x−2 −x−1 + C = Cx − 1 . x3 12. Here, P (x) = 4, Q(x) = x2 e−4x . So, µ(x) = e4x and d 4x e y = x2 dx y = e−4x ⇒ x2 dx = e−4x x3 +C .

2 e−x xdx = − ⇒ ⇒ ydy 1 + y2 ⇒ 2 e−x xdx = − dv u = x2 , v = 1 + y 2 v 2 = − ln |v| + C ⇒ ln 1 + y 2 − e−x = C e−u du = − −e−u is an implicit solution to the given equation. 2 18. Separating variables yields dy = 1 + y2 dy = tan xdx ⇒ 1 + y2 √ Since y(0) = 3, we have tan xdx ⇒ arctan y = − ln | cos x| + C. √ arctan 3 = − ln cos 0 + C = C ⇒ C= π . 3 Therefore, arctan y = − ln | cos x| + π 3 ⇒ y = tan − ln | cos x| + π 3 is the solution to the given initial value problem. 20. Separating variables and integrating, we get (2y + 1)dy = 3x2 + 4x + 2 dx y 2 + y = x3 + 2x2 + 2x + C.

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