# Differential equation substitution

• Sep 3rd 2010, 05:52 PM
acevipa
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?
• Sep 3rd 2010, 06:29 PM
HallsofIvy
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 PM
acevipa
Quote:

Originally Posted by HallsofIvy
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?