Problem source

Let $y,u,v,P$ and $Q$ all be functions of $x$. Show that the substitution $y=uv$ in the differential equation
\[
\frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q
\]
leads to an equation for $\dfrac{\mathrm{d}v}{\mathrm{d}x}$ in terms of $x,Q$ and $u$, provided that $u$ satisfies a suitable first order differential equation. 
Hence or otherwise solve
\[
\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}},
\]
given that $y(1)=0$. For what set of values of $x$ is the solution valid?

Solution source

Suppose $y = uv$ then and suppose $\frac{\d u}{\d x} + P u = 0$ then
\begin{align*}
&& \frac{\d y}{\d x} + Py &= Q \\
&& uv' + u'v + Puv &= Q \\
&& uv' &= Q \\
&& \frac{\d v}{\d x} &= \frac{Q}{u}
\end{align*}

Consider
\begin{align*}
&& 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\
\Rightarrow && \ln u &= 2\ln (1 + x) + C \\
\Rightarrow && u &= A(1+x)^2
\end{align*}
and
\begin{align*}
&&  \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\
\Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\
\Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\
&& 0 &= y(1) \\ 
&&&= \frac23 2^{7/2}+k2^2 \\
\Rightarrow && k &= -\frac{2^{5/2}}{3} \\
\Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2
\end{align*}