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

1. anonymous

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.

2. anonymous

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=

3. anonymous

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. $$e^{ln{|y|}} = e^{-ln|x| - 1/x + ln(4)-1}$$ $$y = \frac{4e^{\frac{-1}{x}-1}}{x}$$