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Math Help - Differential equation substitution

  1. #1
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    Differential equation substitution

    Solve y'-y=xy^2 by letting v=y^{\alpha} for appropriate \alpha.

    How would you go about solving this? I don't really know where to start?
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  2. #2
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    Start by doing exactly what you were told! If v= y^\alpha then y= v^{1/\alpha}. Then y'= \frac{v^{1/\alpha- 1}}{\alpha}v'= \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'.

    Putting that into the equation y'- y= xy^2 gives \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'- v^{1/\alpha}= xv^{\frac{2}{\alpha}. You might now try multiplying the entire equation by v^\alpha and see what you get.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Start by doing exactly what you were told! If v= y^\alpha then y= v^{1/\alpha}. Then y'= \frac{v^{1/\alpha- 1}}{\alpha}v'= \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'.

    Putting that into the equation y'- y= xy^2 gives \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'- v^{1/\alpha}= xv^{\frac{2}{\alpha}. You might now try multiplying the entire equation by v^\alpha and see what you get.
    \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'- v^{1/\alpha}=xv^{\frac{2}{\alpha}

    \frac{v^{1-\alpha}}{\alpha}v'-1=xv^2

    What should I do next?
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