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Thread: Second order non linear inhomogeneous differential equation.. :(

  1. #1
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    Second order non linear inhomogeneous differential equation.. :(

    Solve

    y''= x^3/y

    I tried by using let u=y'

    then I get

    u'=x^3/y .. now something tells me this is not going to be an easy ride home.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Lets write the second order DE as...

    $\displaystyle y^{''} y = x^{3}$ (1)

    ... and suppose that the solution is analytic in $\displaystyle x=0$ so that is...

    $\displaystyle y(x)= a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} + \dots $ (2a)

    $\displaystyle y^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots + n a_{n} x^{n-1} + \dots$ (2b)

    $\displaystyle y^{''} (x) = 2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots + n(n-1) a_{n} x^{n-2} + \dots$ (2c)

    The $\displaystyle a_{n}$ are found inserting the (2a) and (2c) in (1), once we know the 'initial conditions' $\displaystyle y(0)=a_{0}$ and $\displaystyle y^{'} (0) = a_{1}$ ...

    $\displaystyle 2 a_{0} a_{2} =0 \rightarrow a_{2} = 0$

    $\displaystyle 6 a_{0} a_{3} + 2 a_{1} a_{2} = 0 \rightarrow a_{3}=0$

    $\displaystyle 12 a_{0} a_{4} + 6 a_{1} a_{3} + 2a_{2}^{2}= 0 \rightarrow a_{4}=0$

    $\displaystyle 20 a_{0} a_{5} + 12 a_{1} a_{4} + 8a_{2} a_{3}= 1 \rightarrow a_{5}= \frac{a_{0}}{20}$

    $\displaystyle 30 a_{0} a_{6} + 20 a_{1} a_{5} + 12a_{2} a_{4} + 6 a_{3}^{2} = 0 \rightarrow a_{6}= - \frac{a_{1}}{30}$

    ... and so one...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Why assume x=0?

    Why not just start with u=y' ?
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  4. #4
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    Quote Originally Posted by Khonics89 View Post
    Why assume x=0?

    Why not just start with u=y' ?
    Because then you get a single equation with the two unknown functions, u and y.

    And he did NOT "assume x= 0", he assumed the solution was analytic at x= 0.
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