# Differential equation substitution

• Sep 3rd 2010, 06:52 PM
acevipa
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?
• Sep 3rd 2010, 07:29 PM
HallsofIvy
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.
• Sep 3rd 2010, 08:09 PM
acevipa
Quote:

Originally Posted by HallsofIvy
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?