Problem source

\begin{questionparts}
\item In the differential equation 
\[
\frac{1}{y^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y}=\mathrm{e}^{2x}
\]
make the substitution $u=1/y,$ and hence show that the general solution
of the original equation is 
\[
y=\frac{1}{A\mathrm{e}^{x}-\mathrm{e}^{2x}}\,.
\]
\item Use a similar method to solve the equation 
\[
\frac{1}{y^{3}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y^{2}}=\mathrm{e}^{2x}.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item $,$
\begin{align*}
u = 1/y, \frac{\d u}{\d x} = -1/y^2 y': && -\frac{\d u}{\d x} + u &= e^{2x} \\
\Rightarrow && \frac{\d}{\d x} \left (e^{-x}u \right) = -e^{x} \\
\Rightarrow && e^{-x}u &= -e^x+C \\
\Rightarrow && u &= Ce^x-e^{2x} \\
\Rightarrow && y &= \frac{1}{Ce^x-e^{2x}}
\end{align*}

\item $\,$
\begin{align*}
u = 1/y^2, u' = -2/y^3 y': && -2u'+u &= e^{2x} \\
\Rightarrow && \frac{\d}{\d x} \left (e^{-1/2x}u\right) &= -\frac12e^{\frac32x} \\
\Rightarrow && e^{-\frac12x}u &= -\frac13e^{\frac32x}+C \\
\Rightarrow && u &= -\frac13e^{2x}+Ce^{\frac12x} \\
\Rightarrow && y &= \left ( \frac{1}{Ce^{\frac12x}-\frac13e^{2x}}\right)^{\frac12}
\end{align*}

\end{questionparts}