Differential equation substitution

**acevipa** · September 3rd 2010, 05:52 PM

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?

**HallsofIvy** · September 3rd 2010, 06:29 PM

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.

**acevipa** · September 3rd 2010, 07:09 PM

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?

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