# Thread: Diff. Equa. Of 2nd Order

1. ## Diff. Equa. Of 2nd Order

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

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

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

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

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

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

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

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

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

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

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

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

So, the reduced equation is now,

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

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

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

2. For $\displaystyle y''+P y'+Qy=0$, let:

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

in general which reduces it to Normal form:

$\displaystyle v''+Iv=0$

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

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

which is easy to solve.

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