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Math Help - Need help solving (what method to use)

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    Need help solving (what method to use)

    x^2 y'' - (x^2 + 2x)y' + (x + 2) y = 0

    i am supposed to find the 2 independent solutions for this equation but i'm not sure which method to use. i was going to use the methof of solving cauchy-euler equations but i got stuck
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  2. #2
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    Quote Originally Posted by haneek View Post
    x^2 y'' - (x^2 + 2x)y' + (x + 2) y = 0

    i am supposed to find the 2 independent solutions for this equation but i'm not sure which method to use. i was going to use the methof of solving cauchy-euler equations but i got stuck
    By inspection you can see (guess) that one solution is

    y=x

    Now we can use reduction of order

    y=xu(x) \quad y'=u(x)+xu'(x) \quad y''=xu''(x)+2u'(x)

    x^2(xu''(x)+2u'(x))-(x^2+2x)(u(x)+xu'(x) )+(x+2)(xu(x))=0

    Simplifying gives

    x^3u''(x)+(2x^2-x^3-2x^2)u'(x)+(-x^2-2x+x^2+2x)u(x)=0

    x^3(u''(x)-u'(x))=0 \iff \frac{du'}{dx}=u'

    \frac{du'}{u'}=dx \implies \ln|u'|=x+c \implies u'(x)=Ae^{x}

    integrating once more gives

    u(x)=Ae^{x}+B

    Now picking A=1 B=0 gives

    u(x)=e^{x} \implies y=xe^{x}

    So the solution is

    y_c=c_1x+c_2xe^{x}
    Last edited by TheEmptySet; May 19th 2011 at 05:03 AM. Reason: Missing 2nd derivative. Thanks Bugatti79! :)
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    By inspection you can see (guess) that one solution is

    y=x

    Now we can use reduction of order

    y=xu(x) \quad y'=u(x)+xu'(x) \quad y=xu''(x)+2u'(x)()
    Just in case the OP doesnt follow, I think TheEmptySet has a typo in the last equation, ie y should be y''
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    By inspection you can see (guess) that one solution is

    y=x

    Now we can use reduction of order

    y=xu(x) \quad y'=u(x)+xu'(x) \quad y''=xu''(x)+2u'(x)

    x^2(xu''(x)+2u'(x))-(x^2+2x)(u(x)+xu'(x) )+(x+2)(xu(x))=0

    Simplifying gives

    x^3u''(x)+(2x^2-x^3-2x^2)u'(x)+(-x^2-2x+x^2+2x)u(x)=0

    x^3(u''(x)-u'(x))=0 \iff \frac{du'}{dx}=u'

    \frac{du'}{u'}=dx \implies \ln|u'|=x+c \implies u'(x)=Ae^{x}

    integrating once more gives

    u(x)=Ae^{x}+B

    Now picking A=1 B=0 gives

    u(x)=e^{x} \implies y=xe^{x}

    So the solution is

    y_c=c_1x+c_2xe^{x}
    Your welcome. I was meant to ask you why/how you set A=1 and B=0?

    Thanks
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    Quote Originally Posted by bugatti79 View Post
    Your welcome. I was meant to ask you why/how you set A=1 and B=0?

    Thanks
    We are free to pick them because when we form the general solution to the homogenous solution it has the form

    y=c_1x+c_2xu(x)

    So if we left them we would just get new constants in the final solution
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    We are free to pick them because when we form the general solution to the homogenous solution it has the form

    y=c_1x+c_2xu(x)

    So if we left them we would just get new constants in the final solution
    Ok, just one more question. How did you arrive at the general solution to be in the form you have above? I dont see this form anywhere in the thread?
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    Quote Originally Posted by bugatti79 View Post
    Ok, just one more question. How did you arrive at the general solution to be in the form you have above? I dont see this form anywhere in the thread?
    It was mentioned (very briefly) in my first post that I was going to use reduction of order.

    Reduction of order - Wikipedia, the free encyclopedia
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