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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- Sep 3rd 2010, 05:52 PMacevipaDifferential 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? - Sep 3rd 2010, 06:29 PMHallsofIvy
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. - Sep 3rd 2010, 07:09 PMacevipa