function of y second order equation 2

**transgalactic** · November 20th 2010, 12:08 PM

$2(y')^2=y''(y-1)$
$y(1)=2 y'(1)=-1$

again i tried
y'=z
$z'y=z^{2}$
$\frac{{dz}}{dx}=\frac{dz}{dy}\frac{dy}{dx}$
$zz'=dz/dx$

**Ackbeet** · November 20th 2010, 04:29 PM

Try the same trick as on the other post:

$2\,\dfrac{y'}{y-1}=\dfrac{y''}{y'},$ and integrate.

**transgalactic** · November 20th 2010, 11:21 PM

how ?
we dont have a single dy /dx member here
we have many differentiated members on both side

**Prove It** · November 20th 2010, 11:34 PM

You have

$\displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{\frac{dy}{dx}}\,\frac{d^2y}{dx^2}$ .

On the RHS make the substitution $\displaystyle u = \frac{dy}{dx}$ so that $\displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2}$ so that your DE becomes

$\displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{u}\,\frac{du}{dx}$ .

Now integrating both sides gives

$\displaystyle \int{\frac{2}{y-1}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$\displaystyle \int{\frac{2}{y-1}\,dy} = \int{\frac{1}{u}\,du}$

$\displaystyle 2\ln{|y-1|} + C_1 = \ln{|u|} + C_2$

$\displaystyle \ln{|y-1|^2} + C = \ln{|u|}$ , where $\displaystyle C = C_1 - C_2$

$\displaystyle C = \ln{|u|} - \ln{|(y-1)^2|}$

$\displaystyle C = \ln{\left|\frac{u}{(y-1)^2}\right|}$

$\displaystyle C = \ln{\left|\frac{\frac{dy}{dx}}{(y-1)^2}\right|}$

$\displaystyle \pm e^C = \frac{dy}{dx}\,\frac{1}{(y-1)^2}$

$\displaystyle A = \frac{dy}{dx}\,\frac{1}{(y-1)^2}$ , where $\displaystyle A = \pm e^C$ .

You can integrate both sides with respect to $\displaystyle x$ now to find $\displaystyle y$ . You will need to use partial fractions.

**Ackbeet** · November 22nd 2010, 05:20 AM

From my previous post:

$2\ln|y-1|=\ln|y'|+C_{1}.$

On both sides, you see, the derivative of the denominator appears in the numerator. Whenever that happens, you can do a u substitution, and you'll get du/u, which integrates to ln|u|. See how that works?

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