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Math Help - differential equations

  1. #1
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    differential equations

    solve the differential equations
    xy'-2y = x^2
    my work;
    dy/dx = x+ 2y/x

    p(x)= 2/x
    Q(x)= x^2
    integral 2/x = 2lnx
    Ae^2lnx
    y= (integral e^2lnx)(x^2)/(e^2lnx)
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    solve the differential equations
    xy'-2y = x^2
    my work;
    dy/dx = x+ 2y/x

    p(x)= 2/x
    Q(x)= x^2
    integral 2/x = 2lnx
    Ae^2lnx
    y= (integral e^2lnx)(x^2)/(e^2lnx)
    Use the integrating factor method.

    Divide everything by x and the equation becomes

    \frac{dy}{dx} -\frac{2}{x}y = x.

    The integrating factor is e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}.

    So multiply everything by x^{-2} and you get

    x^{-2}\frac{dy}{dx} - 2x^{-3}y = x^{-1}

    \frac{d}{dx}(x^{-2}y) = x^{-1}.

    Can you go from here?
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  3. #3
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    [quote=Prove It;274960]Use the integrating factor method.

    Divide everything by x and the equation becomes

    \frac{dy}{dx} -\frac{2}{x}y = x.

    The integrating factor is e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}.

    how did u get the above step
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    [quote=twilightstr;274978]
    Quote Originally Posted by Prove It View Post
    Use the integrating factor method.

    Divide everything by x and the equation becomes

    \frac{dy}{dx} -\frac{2}{x}y = x.

    The integrating factor is e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}.

    how did u get the above step
    It's a standard technique for first order linear DE's.

    If you have a first order linear DE of the form

    \frac{dy}{dx} + P(x)y = Q(x)

    The integrating factor is I(x) = e^{\int{P(x)\,dx}}.

    Multiply both sides of the DE by the integrating factor and it reduces to

    \frac{d}{dx}(I(x)y) = I(x)Q(x)

    which is usually solvable.
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    Quote Originally Posted by twilightstr View Post
    solve the differential equations
    xy'-2y = x^2
    my work;
    dy/dx = x+ 2y/x

    p(x)= 2/x
    Q(x)= x^2
    integral 2/x = 2lnx
    Ae^2lnx
    y= (integral e^2lnx)(x^2)/(e^2lnx)
    It will help you considerably to recognise that e^{2ln x}= e^{ln x^2}= x^2!
    y= \frac{\int^x t^4 dt}{x^2}

    That is easy to integrate. (I wrote "t" for the dummy variable inside the integral rather than "x" to remind you that you cannot cancel the "x" in the denominator until after you have integrated.)
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    It will help you considerably to recognise that e^{2ln x}= e^{ln x^2}= x^2!
    y= \frac{\int^x t^4 dt}{x^2}

    That is easy to integrate. (I wrote "t" for the dummy variable inside the integral rather than "x" to remind you that you cannot cancel the "x" in the denominator until after you have integrated.)
    You have to divide everything through by x first.

    So P(x) = -\frac{2}{x}, not -2.
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