Problem source

\begin{questionparts}
\item Show that substituting  $y=xv$, where $v$ is a function of $x$,  in the differential equation
\[ xy \frac{\d y}{\d x} +y^2- 2x^2 =0 \quad (x\ne0) \] 
leads to the differential equation
\[ xv\frac{\d v}{\d x} +2v^2 -2=0\,. \]
Hence show that the general solution can be written in the form 
\[ x^2(y^2 -x^2) = C \,,\] where $C$ is a constant. 
\item Find the general solution of the differential equation
\[  y \frac{\d y}{\d x} +6x +5y=0\,  \quad (x\ne0). \]
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& y &= xv \\
&& y' &= v + xv' \\
\Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\
&&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\
\Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\
\\
\Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\
\Rightarrow && \int \frac{v}{1-v^2} \d v  &=2 \ln |x| \\
\Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\
\Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\
\Rightarrow && x^4(1-v^2) &= K \\
\Rightarrow && x^2(x^2-y^2) &= K
\end{align*}

\item $\,$ \begin{align*}
&& 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\
&&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\
\Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\
\Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\
\Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\
&&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\
\Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\
\Rightarrow && -\ln |x| &=  \ln \frac{|v+3|^3}{|v+2|^2} + C \\
\Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\
\Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\
\Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\
\Rightarrow && (y+2x)^2 &= A|y+3x|^3

\end{align*}

\end{questionparts}