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.
 A: you have trouble integrating $$ \frac{10x^2 - 1}{x(10x^2 + 7x + 1)}? $$ this can be broken up what is called the partial fraction decomposition as $$\frac{10x^2 - 1}{x(10x^2 + 7x + 1)} = 
\frac{10x^2 - 1}{x(5x+1)(2x+1)} = \frac{A}{x} + \frac{B}{5x + 1} + \frac C{2x+1} $$
you can find the constants by picking $x$ values carefully. once you know the constants, can you complete it?

$\bf{p.s:}$ i get  $A = -1, B = 5, C = 2$  which gives you $$\int \frac{(10x^2 - 1)}{x(10x^2 + 7x + 1)}\, dx =-\ln| x| + \ln|5x+1| + \ln|2x+1| + C$$
A: 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}
A: What about unqiueness and existaeince of a solution.
Part b) of this question is
Using the existence and uniqueness theorem, what is the largest interval on which a unique solution is guaranteed to exist, if
a. y(1/4)= 5? b. y(-1/4)=5
c. y(-3)=5
How do you go about doing that
EDIT
 I figured it out.  you need to keep the denominator factored _ ie x(2x+1)(5x+1).. so x cant be 0 , -1/5 and -1/2
