Problem source

\begin{questionparts}
\item Show that the gradient at a point $\l x\,, \, y \r$ on the curve 
\[
\l y + 2x \r^3 \l y - 4x \r = c\;,
\]
where $c$ is a constant, is given by
\[
\frac{\d y}{\d x} = \frac{16 x -y}{2y-5x} \;.
\]
\item By considering the derivative with respect to $x$ of 
$\l y + ax \r^n \l y + bx \r\,$, or otherwise, 
find the general solution of the differential equation
\[
\frac{\mathrm{d}y}{ \mathrm{d}x} = \frac{10x - 4y}{  3x - y}\;.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& c &= \left (y+2x \right)^3\left (y-4x \right) \\
\Rightarrow && 0 &= 3\left (y+2x \right)^2\left (y-4x \right)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right)^3 \left ( \frac{\d y}{\d x} - 4 \right) \\
\Rightarrow && 0 &= 3(y-4x)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right) \left ( \frac{\d y}{\d x} - 4 \right) \\
\Rightarrow &&&= \frac{\d y}{\d x} \left (3(y-4x) + (y+2x) \right) + 6(y-4x)-4(y+2x) \\
&&&= \frac{\d y}{\d x} \left ( 4y-10x\right) + 2y-32x \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{16x-y}{2y-5x}
\end{align*}
\item \begin{align*}
&& c &= \left ( y + ax \right)^n \left ( y + bx \right) \\
\Rightarrow && 0 &= n\left ( y + ax \right)^{n-1} \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right)^{n} \left ( \frac{\d y}{ \d x}+b \right) \\
\Rightarrow && 0 &= n \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right) \left ( \frac{\d y}{ \d x}+b \right) \\
&&&= \frac{\d y}{\d x} \left ( (n+1)y  + (nb+a)x \right) + an(y+bx) + by+bax \\
&&&= \frac{\d y}{\d x} \left ( (n+1)y  + (nb+a)x \right) + (an+b)y+ab(n+1)x \\
\Rightarrow && \frac{\d y}{\d x} &= -\frac{(an+b)y+ab(n+1)x}{(n+1)y+(nb+a)x}
\end{align*}

We must have $ab = 10, a+b = -7$ so say $a=-5,b=-2,n=2$ and we have

$(y-5x)^2(y-2) = c$ is our general solution to the differential equation



\end{questionparts}