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Math Help - Differential equations

  1. #1
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    Differential equations

    Solve:
    x^2\frac{d^2 y}{dx^2} + 2x\frac{dy}{dx} = \frac{1}{X^2}
    and i am very stuck .
    Can i integrate the whole thing? But im unsure as to what you get because of the coefficients...
    Then Dividing by the coefficients doesnt offer me much help either lol =(
    Thanks.
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Solve:
    x^2\frac{d^2 y}{dx^2} + 2x\frac{dy}{dx} = \frac{1}{X^2}
    and i am very stuck .
    Can i integrate the whole thing? But im unsure as to what you get because of the coefficients...
    Then Dividing by the coefficients doesnt offer me much help either lol =(
    Thanks.
    I'm just starting my study of differential equations; however, I think that I can help. This equation looks like the derivative of a product. Y= -1/2*x^-2 - c*x^-1 +c
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  3. #3
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    Quote Originally Posted by Bilbo Baggins View Post
    I'm just starting my study of differential equations; however, I think that I can help. This equation looks like the derivative of a product. Y= -1/2*x^-2 - c*x^-1 +c
    Hmm, did you have to find that by guessing?
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  4. #4
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    Quote Originally Posted by AshleyT View Post
    Solve:
    x^2\frac{d^2 y}{dx^2} + 2x\frac{dy}{dx} = \frac{1}{X^2}
    and i am very stuck .
    Can i integrate the whole thing? But im unsure as to what you get because of the coefficients...
    Then Dividing by the coefficients doesnt offer me much help either lol =(
    Thanks.
    You'll notice there's no y is this problem so it you let

    u = \frac{dy}{dx},\;\;\; \text{then}\;\;\; \frac{du}{dx} = \frac{d^2 y}{dx^2}

    and your problem becomes

    x^2 \frac{du}{dx} + 2x u = \frac{1}{x^2} linear!
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  5. #5
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    Quote Originally Posted by AshleyT View Post
    Hmm, did you have to find that by guessing?
    No, not at all. I just saw it. The derivative of a product is pretty conspicuous.
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  6. #6
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    Quote Originally Posted by Bilbo Baggins View Post
    No, not at all. I just saw it. The derivative of a product is pretty conspicuous.
    Oh ok, thanks =)
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