Find an explicit solution of the given initial-value problem.
x^(2) y' = y − xy, y(−1) = −4

Question

Find an explicit solution of the given initial-value problem.
x^(2) y' = y − xy,  y(−1) = −4

anonymous · Answer

You can do separation of variables to solve this:

$x^{2}y' = y-xy$

$x^{2}y' = y(1-x)$

$\frac{dy}{y} = \frac{(1-x)}{x^{2}}dx$

Then you can just integrate both sides and go from there.

anonymous · Answer

Thank you I've tried that and after doing the integration i get ln(y)= -x^(-1)-ln(x)+C. And I know I have to plug in the initial conditions to solve for C. Which I got as C=ln(4)-1 But they want the final answer as y=

anonymous · Answer

Okay, so that means you would have 

$ln|y|= -ln|x|-\frac{1}{x}+ ln(4) - 1$

Just raise both sides as the exponent of e then and simplify. 

$e^{ln{|y|}} = e^{-ln|x| - 1/x + ln(4)-1}$

$y = \frac{4e^{\frac{-1}{x}-1}}{x}$