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Thread: Differential equation substitution

  1. #1
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    Differential equation substitution

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

    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 $\displaystyle v= y^\alpha$ then $\displaystyle y= v^{1/\alpha}$. Then $\displaystyle y'= \frac{v^{1/\alpha- 1}}{\alpha}v'= \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'$.

    Putting that into the equation $\displaystyle y'- y= xy^2$ gives $\displaystyle \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'- v^{1/\alpha}= xv^{\frac{2}{\alpha}$. You might now try multiplying the entire equation by $\displaystyle v^\alpha$ 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 $\displaystyle v= y^\alpha$ then $\displaystyle y= v^{1/\alpha}$. Then $\displaystyle y'= \frac{v^{1/\alpha- 1}}{\alpha}v'= \frac{v^{(1-\alpha)/\alpha}}{\alpha}v'$.

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

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

    What should I do next?
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