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Thread: Diff. Equa. Of 2nd Order

  1. #1
    Member kjchauhan's Avatar
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    Diff. Equa. Of 2nd Order

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

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

    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..
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  2. #2
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    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.
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  3. #3
    Member kjchauhan's Avatar
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    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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