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Math Help - Non-linear DE #2

  1. #1
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    Non-linear DE #2

    (1-x)y'' + xy' - y = 0

    i have a midterm tomorrow and i'm having a lot of trouble with these types of DEs
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  2. #2
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    Quote Originally Posted by haneek View Post
    (1-x)y'' + xy' - y = 0

    i have a midterm tomorrow and i'm having a lot of trouble with these types of DEs
    once again can you "guess" as solution? It is a simple solution.
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    guess what?
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  4. #4
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    Quote Originally Posted by haneek View Post
    guess what?
    Did you read what was posted at your other thread! This is the exact same type of problem. If you can guess one solution you can use reduction of order to find the complete soluton set.
    Last edited by mr fantastic; May 18th 2011 at 07:33 PM.
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    yes i did! is one of the solutions (x+1)?
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    Quote Originally Posted by haneek View Post
    yes i did! is one of the solutions (x+1)?
    No, remember that you can check if it is by plugging it into the ode.,
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    ok so i continued solving the problem and i choose that one of the solutions is x and i plugged it and it works. then i got the derivates of y = u(x) x
    and i reached this step

    u''[x+x^2] + u'[x^2-2x+2] = 0

    but now i'm stuck i dont know how to continue to find y2
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by haneek View Post
    ok so i continued solving the problem and i choose that one of the solutions is x and i plugged it and it works. then i got the derivates of y = u(x) x
    and i reached this step

    u''[x+x^2] + u'[x^2-2x+2] = 0

    but now i'm stuck i dont know how to continue to find y2
    Im also curious how one would tackle this although I get u''[x-x^2] + u'[x^2-2x+2] = 0

    Would this be a correct attempt -

    \dispalystyle \int\frac{u''(x)}{u'(x)}dx=\int\frac{x^2-2x+2}{x-x^2}dx
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  9. #9
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    By the way, you titled this thread "non-linear DE #2". You should understand that this is NOT a non-linear differential equation- it is a perfectly good linear equation.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    By the way, you titled this thread "non-linear DE #2". You should understand that this is NOT a non-linear differential equation- it is a perfectly good linear equation.
    I realise its not non-linear but I am not the OP. I am just also interested in the solution

    Would that be a correct approach?
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  11. #11
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    Quote Originally Posted by bugatti79 View Post
    Im also curious how one would tackle this although I get u''[x-x^2] + u'[x^2-2x+2] = 0

    Would this be a correct attempt -

    \dispalystyle \int\frac{u''(x)}{u'(x)}dx=\int\frac{x^2-2x+2}{x-x^2}dx
    Yes it just takes some work. Here we go!

    \ln|u'(x)|=x+\ln|x-1|-2\ln|x|+\ln(c) \iff u'(x)=c\left( \frac{1}{x}e^{x}-\frac{1}{x^2}e^{x}\right)

    Now if we integrate by parts with

    u=e^{x} \implies du=e^x \text{ and } dv=\frac{1}{x^2} \implies v= -\frac{1}{x}

    This gives

    \frac{1}{c}u(x)=\int \frac{e^x}{x}dx-\left[ -\frac{e^x}{x}+\int \frac{e^x}{x}dx\right]=\frac{e^x}{x}+d

    So pick c=1 d=0 and we get the 2nd solution is

    y=xu(x)=x\left( \frac{e^x}{x}\right)=e^x

    u=c_1x+c_2e^x
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  12. #12
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Yes it just takes some work. Here we go!

    \ln|u'(x)|=x+\ln|x-1|-2\ln|x|+\ln(c) \iff u'(x)=c\left( \frac{1}{x}e^{x}-\frac{1}{x^2}e^{x}\right)

    Now if we integrate by parts with

    u=e^{x} \implies du=e^x \text{ and } dv=\frac{1}{x^2} \implies v= -\frac{1}{x}

    This gives

    \frac{1}{c}u(x)=\int \frac{e^x}{x}dx-\left[ -\frac{e^x}{x}+\int \frac{e^x}{x}dx\right]=\frac{e^x}{x}+d

    So pick c=1 d=0 and we get the 2nd solution is

    y=xu(x)=x\left( \frac{e^x}{x}\right)=e^x

    u=c_1x+c_2e^x
    Hmmm, the same here, why/how c=1, d=0? :-)
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