Answer
$y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+1)$
Work Step by Step
The standard form of the given equation is:
$y'+\dfrac{2}{x+1} y=\dfrac{x}{x+1}$ ....(1)
The integrating factor is: $e^{\int (\dfrac{2}{x+1}) dx}=(x+1)^2$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [(x+1)^2y]' =\int x(x+1) dx \implies (x+1)^2 y=\int (x^2+x) dx$
or, $y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+c)$
Applying the initial conditions, we get $c=1$
Thus, we have
$y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+1)$
