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Solve second order diff eq with substitution

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve:

    [tex] xy''+2y'=12x^{2} [/tex]

    with

    [tex] u=y' [/tex]

    2. Relevant equations

    if you have:

    [tex] y'+P(x)y=Q(x) [/tex]

    then your integrating factor is:

    [tex] I(x)=e^{\int P(x) dx} [/tex]

    3. The attempt at a solution

    The only reason I was able to solve this is because I stumbled upon a similar post here but instead of pulling up a thread from over a year ago i decided to post my specific question.

    So if:

    [tex] xy''+2y'=12x^{2} \hookrightarrow y''+\frac{2y'}{x}=12x[/tex]

    and

    [tex] y'=u \hookrightarrow y''=u\frac{du}{dy} [/tex]

    I do not understand why y'' is equal to this. The way I read y'' is d2y/dx2 so why is it u*du/dy as the derivative on the right side and not d/dx? And! why is the derivative of u with respect to y u*du/dy and not just du/dy.

    Maybe I have just forgotten something simple from differential calculus but I can't make sense of it on my own. I am not claiming that what is written in TeX above is wrong because the problem seems to work out, I just don't know why that is the way to do it.

    Anyway, if we make the substitution:

    [tex] u\frac{du}{dy}+\frac{2u}{x}=12x [/tex]

    then

    [tex] I(x)=e^{\int \frac {2}{x} dx} = e^{2 ln|x|}=x^{2} [/tex]

    then multiply both sides by I(x):

    [tex] x^{2}u\frac{du}{dy}+2xu=12x^{3} [/tex]

    which is:

    [tex] \frac {d}{dx} (ux^{2})=12x^{3} [/tex]

    [tex] ux^{2}=\int 12x^{3} dx =3x^{4}+C_{1} [/tex]

    [tex] u=3x^{2}+ \frac{C_{1}}{x^{2}} [/tex]

    [tex] \frac {dy}{dx}=3x^{2} + \frac {C_{1}}{x^{2}} [/tex]

    [tex] y= \int 3x^{2}+\frac{C_{1}}{x^{2}}dx = x^{3}-\frac{C_{1}}{x}+C_{2} [/tex]

    if that is right please tell me why. I just followed by example from someone elses work.
     
  2. jcsd
  3. Jun 28, 2011 #2

    vela

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    That's just the chain rule.
    [tex]y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u\frac{du}{dy}[/tex]
    I'm not sure why you'd do it that way, though. You could simply write
    [tex]x u' + 2 u = 12x^2[/tex]
    and solve it with the same integrating factor without the unnecessary complication of invoking the chain rule.
     
  4. Jun 28, 2011 #3
    oh haha alright. I guess I was confused by the substitution of y'=u again for some reason.
    Thanks for the reminder on that!

    So I assume this is a correct solution? It's an even problem so I can't look it up.
     
  5. Jun 28, 2011 #4

    vela

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    I didn't see anything obviously wrong, but I didn't look that closely either. You can always check your answer by plugging it back into the original differential equation to see if it's satisfied.
     
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