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Thread: Checking differential equation

  1. #1
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    Checking differential equation

    Can someone check my working?

    Solve $\displaystyle xy'=2x^2y+y\ln y$ by letting $\displaystyle \ln y=v$

    $\displaystyle \ln y=v\Rightarrow y=e^v$

    $\displaystyle \dfrac{dy}{dx}=\dfrac{dy}{dv}\cdot\dfrac{dv}{dx}$

    $\displaystyle \dfrac{dy}{dx}=e^v \dfrac{dv}{dx}$

    $\displaystyle xe^v \dfrac{dv}{dx}=2x^2e^v+ve^v$

    Divide by $\displaystyle e^v$

    $\displaystyle x \dfrac{dv}{dx}=2x^2+v$

    $\displaystyle \dfrac{dv}{dx}=2x+\dfrac{v}{x}$

    $\displaystyle \dfrac{dv}{dx}-\dfrac{v}{x}=2x$

    $\displaystyle I=e^{\int{-\frac{1}{x}} \ dx}=\dfrac{1}{x}$

    $\displaystyle \dfrac{1}{x} \dfrac{dv}{dx}-\dfrac{v}{x^2}=2$

    $\displaystyle \left[\dfrac{1}{x}v\right]'=2$

    $\displaystyle \dfrac{1}{x}v=2x+C$

    $\displaystyle v=2x^2+Cx$

    $\displaystyle \ln y =2x^2+Cx$

    $\displaystyle \Rightarrow y=e^{2x^2+Cx}$
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  2. #2
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    It all looks good to me...
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