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Math Help - x^2 y'' - 5x y' - 7y = x^3

  1. #1
    s3a
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    x^2 y'' - 5x y' - 7y = x^3

    The question asks me to solve x^2 y'' - 5x y' - 7y = x^3 as function of x. In other words, it's asking for y(x). I am given the initial conditions y(1) = 10 and y'(1) = 8.

    Someone who wasn't very sure told me to use characteristic equations and the method of undetermined coefficients but is that correct? If so, how do I set it up mechanically? I only know how to use characteristic equations when there are no x variables in the differential equation. I am really, really stuck. I only divided both sides by x^2.

    The answer is:
    http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3;+y(1)+%3D+10,+y'(1)+%3D+8

    If someone could show me how to set up the problem (tedious steps aren't necessary), I would greatly appreciate it!

    Thanks in advance!
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    Re: x^2 y'' - 5x y' - 7y = x^3

    Quote Originally Posted by s3a View Post
    The question asks me to solve x^2 y'' - 5x y' - 7y = x^3 as function of x. In other words, it's asking for y(x). I am given the initial conditions y(1) = 10 and y'(1) = 8.

    Someone who wasn't very sure told me to use characteristic equations and the method of undetermined coefficients but is that correct? If so, how do I set it up mechanically? I only know how to use characteristic equations when there are no x variables in the differential equation. I am really, really stuck. I only divided both sides by x^2.

    The answer is:
    http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3;+y(1)+%3D+10,+y'(1)+%3D+8

    If someone could show me how to set up the problem (tedious steps aren't necessary), I would greatly appreciate it!

    Thanks in advance!
    You can use the substitution

    y=f(u) where u=ln(x) Then by the chain rule you get

    \frac{dy}{dx}=\frac{df}{du}\frac{du}{dx}=\frac{1}{  x}f'

    Now if you take one more derivative you get

    \frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{1}{x}f'\right)=-\frac{1}{x^2}f'+\frac{1}{x}f''\cdot \frac{1}{x}=\frac{1}{x^2}(-f'+f'')

    Now just sub this into the equation and watch the magic happen.
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    s3a
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    Re: x^2 y'' - 5x y' - 7y = x^3

    Sorry for the dumb question but, is this process just a convenient trick or is it how I am supposed to solve most differential equations with x variables in them? Also, is there a name for this process other than just "substitution?"
    Last edited by s3a; November 29th 2011 at 12:11 PM.
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    s3a
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    Re: x^2 y'' - 5x y' - 7y = x^3

    Also, I think I almost get it but I get the wrong answer. I'm stuck at evaluating the new initial conditions as well so it would be great if you could help me with that too but I have the answer without the initial conditions plugged in (http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3) and it differs from the answer I get.

    I'm attaching my work.
    Attached Thumbnails Attached Thumbnails x^2 y'' - 5x y' - 7y = x^3-n6.jpg  
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    Re: x^2 y'' - 5x y' - 7y = x^3

    The equation is of the type Cauchy- Euler. A Google search will turn up alot of info.

    I was a little careless with my notation above and I think it has confused you. When we do this substitution we are chainging the independant variable from x to u. So we need to sub out the x's in the original equation. Using the work above you should get the equation.

    x^2\left[ \frac{1}{x^2}\left( -f'(u)+f''(u)\right)\right]-5x\left( \frac{1}{x}f'(u)\right)-7f(u)=\left( e^{u}\right)^3

    Simplifying will reduce out all of the x's and gives the equation.

    Note: here the primes represent differentation with respect to u.

    f''(u)-6'f(u)-7f(u)=e^{3u}

    I hope this clears it up a bit.

    TES
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