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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can do separation of variables to solve this: \(x^{2}y' = yxy\) \(x^{2}y' = y(1x)\) \(\frac{dy}{y} = \frac{(1x)}{x^{2}}dx\) Then you can just integrate both sides and go from there.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so that means you would have \(lny= lnx\frac{1}{x}+ ln(4)  1\) Just raise both sides as the exponent of e then and simplify. \(e^{ln{y}} = e^{lnx  1/x + ln(4)1}\) \(y = \frac{4e^{\frac{1}{x}1}}{x}\)
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