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Thread: Find the general solution

  1. #1
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    Find the general solution

    Find the general solution to the first order linear diff. equation

    dy/dx + 3xy = -2x
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  2. #2
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    Hello, matty888!

    Find the general solution of: .$\displaystyle
    \frac{dy}{dx} + 3xy \:=\: -2x$

    Integrating factor: .$\displaystyle I \;=\;e^{\int3x\,dx} \;=\;e^{\frac{3}{2}x^2} $

    Multiply: .$\displaystyle e^{\frac{3}{2}x^2}\frac{dy}{dx} + 3xye^{\frac{3}{2}x^2} \;=\;-2xe^{\frac{3}{2}x^2} $

    We have: .$\displaystyle \frac{d}{dx}\left(e^{\frac{3}{2}x^2}y\right) \;=\;-2xe^{\frac{3}{2}x^2} $

    Integrate: .$\displaystyle e^{\frac{3}{2}x^2}y \;=\;-2\int e^{\frac{3}{2}x^2}(x\,dx) $


    On the right side:

    . . Let: $\displaystyle u \:=\:\tfrac{3}{2}x^2 \quad\Rightarrow\quad du \:=\:3x\,dx \quad\Rightarrow\quad x\,dx \:=\:\tfrac{1}{3}du $

    . . Substitute: .$\displaystyle -2\int e^u\left(\tfrac{1}{3}du\right) \;=\;-\tfrac{2}{3}\int e^u\,du \;=\;-\tfrac{2}{3}e^u + C$

    . . Back-substitute: .$\displaystyle -\tfrac{2}{3}e^{\frac{3}{2}x^2} + C$


    The equation becomes: .$\displaystyle e^{\frac{3}{2}x^2}y \;=\;-\tfrac{2}{3}e^{\frac{3}{2}x^2} + C $


    Therefore: .$\displaystyle \boxed{y \;=\;-\frac{2}{3} + Ce^{-\frac{3}{2}x^2}} $

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  3. #3
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    Thanks Soroban

    Hi Soroban,
    Great answer,very clear & helpful,many thanks!!
    matty888
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