Diff. Equa. Of 2nd Order

**kjchauhan** · December 9th 2009, 02:49 PM

Please help me to Solve these Diff. Equ.s by transforming an independent variable: (Where I'm wrong?)

$\frac{d^2y}{dx^2}+\frac{2}{x}\frac{dy}{dx}-n^2y=0$

Take $z=\int e^{-\int Pdx}\;dx$ to make $P_1=0$

Here $P=\frac{2}{x}$

$\therefore z=\int e^{-\int \frac{2}{x}\,dx}\;dx$

$\therefore z=\int e^{-2\log x}\, dx$

$\therefore z= \int e^{\log x^{-2}}\, dx$

$\therefore z=\int x^{-2} \, dx$

$\therefore z=\frac{-1}{x}$

Now,
$\frac{dz}{dx}=x^{-2}$

so,
$Q_1=\frac{Q}{{\left( \frac{dz}{dx}\right)}^2}$

$\therefore Q_1=\frac{-n^2}{{\left( x^{-2} \right)}^2}$

$\therefore Q_1=\frac{-n^2}{z^4}$

So, the reduced equation is now,

$\frac{d^2y}{dz^2}+Q_1y=0$

$\frac {d^2y}{dz^2}-\frac{n^2}{z^4}y=0$

Ok, now I do not work out... pl help..

**shawsend** · December 10th 2009, 03:30 AM

For $y''+P y'+Qy=0$ , let:

$y=v\;\text{exp}\left\{-1/2 \int P dx\right\}$

in general which reduces it to Normal form:

$v''+Iv=0$

where $I=Q-1/2 P'-1/4 P^2$

I get: $v''-n^2 v=0$

which is easy to solve.

**kjchauhan** · December 10th 2009, 03:45 AM

Thanks..
I got it, but as mentioned in book, the method can not applied..(Method of transforming independent variable.)
Thanks for valuable suggestion..

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