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Math Help - Differential Equations

  1. #1
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    Differential Equations

    Would someone be able to point me in the right direction with this problem?

    Find the general solution of the differential equation.

    x^2*y'+x(x+2)y=e^x

    This is what I have so far:

    x^2(dy/dx)+x(x+2)y=e^x

    x^2(dy/dx)+x^2y+2xy=e^x

    dy/dx+y+(2y)/x=e^2/x^2


    e^(ln(x^2))=x^2

    d/dx(x^2y+(2yx^2)/x))=e^x

    1/3*x^3y+2yx^2=e^x

    I'm not sure where to go from here or if I am even doing this correctly. Any help with the steps of this problem would be great. Thanks!
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  2. #2
    Senior Member Peritus's Avatar
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    \begin{gathered}<br />
  y' + \left( {1 + \frac{2}<br />
{x}} \right)y = \frac{{e^x }}<br />
{{x^2 }} \hfill \\<br />
  \mu  = e^{\int {1 + \frac{2}<br />
{x}} dx}  = e^{x + 2\ln x}  = x^2 e^x  \hfill \\ <br />
\end{gathered}

    <br />
\begin{gathered}<br />
  x^2 e^x y' + x(x + 2)e^x y = e^{2x}  \hfill \\<br />
  x^2 e^x y = \int {e^{2x} dx}  \hfill \\ <br />
\end{gathered} <br />

    ...
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  3. #3
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    Quote Originally Posted by yellowrose View Post
    Would someone be able to point me in the right direction with this problem?

    Find the general solution of the differential equation.

    x^2*y'+x(x+2)y=e^x

    This is what I have so far:

    x^2(dy/dx)+x(x+2)y=e^x

    x^2(dy/dx)+x^2y+2xy=e^x

    dy/dx+y+(2y)/x=e^2/x^2


    e^(ln(x^2))=x^2

    d/dx(x^2y+(2yx^2)/x))=e^x

    1/3*x^3y+2yx^2=e^x

    I'm not sure where to go from here or if I am even doing this correctly. Any help with the steps of this problem would be great. Thanks!
    I'd divide through by x^2 and re-wrte it in the form

    \frac{d}{dx} = \left( 1 + \frac{2}{x} \right) y = \frac{e^x}{x^2}.

    Then I'd use the integrating factor method.

    Note that the integrating factor here is e^{\int 1 + \frac{2}{x} \, dx} = e^{x + 2 \ln x} = e^x e^{2 \ln x} = e^x e^{\ln x^2} = x^2 e^x.

    It should be all blue sky from here .....
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  4. #4
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    Quote Originally Posted by Peritus View Post
    \begin{gathered}<br />
  y' + \left( {1 + \frac{2}<br />
{x}} \right)y = \frac{{e^x }}<br />
{{x^2 }} \hfill \\<br />
  \mu  = e^{\int {1 + \frac{2}<br />
{x}} dx}  = e^{x + 2\ln x}  = x^2 e^x  \hfill \\ <br />
\end{gathered}

    <br />
\begin{gathered}<br />
  x^2 e^x y' + x(x + 2)e^x y = e^{2x}  \hfill \\<br />
  x^2 e^x y = \int {e^{2x} dx}  \hfill \\ <br />
\end{gathered} <br />

    ...
    Following your answer I get: x^2*e^x*y=(1/2)e^2x

    This is where I solve for y right?

    Doing this I get: y=((1/2)e^2x)/(x^2*e^x))
    Which equals: y=1/(2x^2)*e^x
    Where do I get the second part from? I should have:
    y=1/(2x^2)*e^x[B]+(c/x^2)*e^-x
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  5. #5
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    Quote Originally Posted by yellowrose View Post
    Following your answer I get: x^2*e^x*y=(1/2)e^2x + C Mr F adds the + C. The arbitrary constant of integration - so often overlooked - is essential to the solution!

    This is where I solve for y right? Mr F says: Yep, but you'll only get half the answer without the + C.

    Doing this I get: y=((1/2)e^2x)/(x^2*e^x))
    Which equals: y=1/(2x^2)*e^x
    Where do I get the second part from? I should have:

    y=1/(2x^2)*e^x[B]+(c/x^2)*e^-x Mr F says: Hate to say I told you so . You only got half the answer because you forgot the arbitrary constant of integration.
    So you should have x^2 e^x y = \frac{1}{2} e^{2x} + C.

    Now solve for y and you'll get the second part!

    Indefinite integrals ALWAYS have an arbitrary constant attached! Do you C? (ha ha)
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