To determine The general solution of the nonlinear ordinary differential equation 2 x y y ′ + ( x − 1 ) y 2 = x 2 e x with the initial condition y ( 0 ) = π 2 . Explanation The given nonlinear ordinary differential equation is 2 x y y ′ + ( x − 1 ) y 2 = x 2 e x . The ordinary differential equation is rewritten as follows. 2 x y y ′ + ( x − 1 ) y 2 = x 2 e x y y ′ + x − 1 2 x ⋅ y 2 = x e x 2 Consider z = y 2 and the derivative is obtained as follows. z ′ = 2 y 2 − 1 ⋅ y ′ y y ′ = z ′ 2 z ′ 2 + x − 1 2 x ⋅ z = x e x 2 z ′ + x − 1 x ⋅ z = x e x The above differential equation is of the form x ′ + p x = r , where p = x − 1 x and r = x e x . The general solution is given by, x ( y ) = e − h ( c + ∫ r e h d y ) ... ( 1 ) The integrating factor h is ∫ p d x and it is calculated as follows

10th Edition

ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 1.5, Problem 28P

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The general solution of the nonlinear ordinary differential equation 2xyy′+(x−1)y2=x2ex with the initial condition y(0)=π2.

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Chapter 1.5, Problem 27P

Chapter 1.5, Problem 29P

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