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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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. \(e^{ln{|y|}} = e^{-ln|x| - 1/x + ln(4)-1}\) \(y = \frac{4e^{\frac{-1}{x}-1}}{x}\)

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