Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.1 Differential Equations Everywhere - Problems - Page 11: 6

Answer

As given below

Work Step by Step

We are given: $y(x)=\frac{x}{x+1}$ $\frac{dy}{dx}=\frac{1}{(1+x)^2}$ $y''(x)=\frac{-2}{(x+1)^3}$ The differential equation: $y+\frac{d^2y}{dx^2}=\frac{dy}{dx}+\frac{x^3+2x^2-3}{(1+x)^3}$ $\frac{x}{x+1}-\frac{2}{(x+1)^3}=\frac{1}{(1+x)^2}+\frac{x^3+2x^2-3}{(1+x)^3}$ $\frac{x(x+1)^2-2}{(x+1)^3}=\frac{(x+1)+x^3+2x^2-3}{(1+x)^3}$ $\frac{x^3+2x^2+x-2}{(x+1)^3}=\frac{x^3+2x^2+x-2}{(1+x)^3}$ Both sides are equal. So the differential equation does exist: $y+\frac{d^2y}{dx^2}=\frac{dy}{dx}+\frac{x^3+2x^2-3}{(1+x)^3}$ and $y(x)=\frac{x}{x+1}$ is a solution to the given differential equation.
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