# z=y' solution

• Dec 8th 2010, 03:04 AM
transgalactic
z=y' solution
xy''+2y'=x(y')^2
z=y'
xz'+2z=xz^2
i cant solve it
• Dec 8th 2010, 03:44 AM
Prove It
$\displaystyle \displaystyle x\,\frac{dz}{dx} + 2z = x\,z^2$

$\displaystyle \displaystyle \frac{dz}{dx} + 2x^{-1}z = z^2$.

This is a Bernoulli Equation, so make the substitution $\displaystyle \displaystyle v = z^{1 - 2} = z^{-1}$.

Then $\displaystyle \displaystyle z = v^{-1}$, so

$\displaystyle \displaystyle \frac{dz}{dx} = \frac{d}{dx}(v^{-1})$

$\displaystyle \displaystyle \frac{dz}{dx} = \frac{d}{dv}(v^{-1})\,\frac{dv}{dx}$

$\displaystyle \displaystyle \frac{dz}{dx} = -v^{-2}\,\frac{dv}{dx}$.

Substituting into the DE $\displaystyle \displaystyle \frac{dz}{dx} + 2x^{-1}z = z^2$ gives

$\displaystyle \displaystyle -v^{-2}\,\frac{dv}{dx} + 2x^{-1}v^{-1} = (v^{-1})^2$

$\displaystyle \displaystyle -v^{-2}\,\frac{dv}{dx} + 2x^{-1}v^{-1} = v^{-2}$

$\displaystyle \displaystyle \frac{dv}{dx} - 2x^{-1}v = -1$.

Now this is First Order Linear, so use the Integrating Factor method.

The Integrating Factor is $\displaystyle \displaystyle e^{\int{-2x^{-1}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}$.

Multiplying through by the Integrating Factor gives

$\displaystyle \displaystyle x^{-2}\,\frac{dv}{dx} - 2x^{-3}v = -x^{-2}$

$\displaystyle \displaystyle \frac{d}{dx}(x^{-2}v) = -x^{-2}$

$\displaystyle \displaystyle x^{-2}v = \int{-x^{-2}\,dx}$

$\displaystyle \displaystyle x^{-2}v = x^{-1} + C$

$\displaystyle \displaystyle v = x + Cx^2$.

Since $\displaystyle \displaystyle z = v^{-1}$

$\displaystyle \displaystyle z = (x + Cx^2)^{-1} = \frac{1}{x + Cx^2} = \frac{1}{x(1 + Cx)}$.

And finally $\displaystyle \displaystyle y = \int{z\,dx} = \int{\frac{1}{x(1+Cx)}\,dx}$. You can do this using Partial Fractions.