<s>[INST] differential equation; how to find the general solution. how do i get the general solution of $$\frac{dy}{dx} = \frac{(10x^2-1)y}{  x(10x^2+7x+1)}$$  please? i have been trying for a while now. I have been doing it as a separable DE and I get $$\frac {dy}y = \frac{(10x^2-1)\, dx}{x(10x^2+7x+1)} $$I can not seem to carry on.
 [/INST] The solution is as follows
\begin{align}
\frac{dy}{dx} = \frac{(10x^2-1)y}{x(10x^2+7x+1)}
\end{align}
can be seen to be of the form
\begin{align}
\frac{dy}{dx} = y \left( \frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x} \right)
\end{align}
or
\begin{align}
\frac{dy}{y} = \left(\frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x} \right) dx
\end{align} 
for which
\begin{align}
\ln(y) = \ln(2x+1) + \ln(5x+1) - \ln(x) + \ln(c_{0}) = \ln\left( \frac{c_{0} (2x+1)(5x+1)}{x} \right) 
\end{align}
and the general solution for $y(x)$ becomes
\begin{align}
y(x) = \frac{c_{0} (2x+1)(5x+1)}{x}.
\end{align}
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