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.
