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Math Help - dont know how to start on thee 2 differential equations

  1. #1
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    dont know how to start on thee 2 differential equations

    Please help, I need to solve this differential equation:

    dy/dx + xy = x^3 , f(0)=1

    AND

    (e^(xy))(1+xy)dx + (x^2)(e^(xy))dy = 0

    I cannot seem to separate the variables in the first one and can't find a homogeneous solution and the second one I just had a mental blank...
    please show me how to solve step by step?
    thanks
    Last edited by zen-x; May 26th 2009 at 01:20 PM. Reason: maybe a reply with more detail
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  2. #2
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    Quote Originally Posted by zen-x View Post

    dy/dx + xy = x^3 , f(0)=1
    You need to use the integrating factor method on this equation as it is not separable.

    for an equation in the from y'+f(x)y = g(x)

    your integrating factor is found as follows

    I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}

    multiplying this through the entire equation gives

    e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}

    then using the product rule, LHS changes to be

    (e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}

    now integrate both sides

    e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx

    gives y to be

    y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}

    you will have to finish the problem by using integration by parts on the numerator of the RHS.

    use f(0) = 1 to determine your constant of integration afterwards.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    You need to use the integrating factor method on this equation as it is not separable.

    for an equation in the from y'+f(x)y = g(x)

    your integrating factor is found as follows

    I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}

    multiplying this through the entire equation gives

    e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}

    then using the product rule, LHS changes to be

    (e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}

    now integrate both sides

    e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx

    gives y to be

    y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}

    you will have to finish the problem by using integration by parts on the numerator of the RHS.
    First write that as \int (x^2e^{\frac{x^2}{2}})(x dx) and let u= x^2. Then use integration by parts.

    use f(0) = 1 to determine your constant of integration afterwards.
    Last edited by HallsofIvy; June 11th 2009 at 11:40 AM.
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  4. #4
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    Quote Originally Posted by zen-x View Post
    (e^(xy))(1+xy)dx + (x^2)(e^(xy))dy = 0
    you need to do a little algebra on this one to put it in the correct form.

    e^{xy}(1+xy)dx + x^2e^{xy}dy = 0

    e^{xy}(1+xy)dx = -x^2e^{xy}dy

    e^{xy}(1+xy)dx = -x^2e^{xy}dy

    divide both sides by e^{xy}

    (1+xy)dx = -x^2dy

    \frac{(1+xy)}{-x^2} = \frac{dy}{dx}

     \frac{dy}{dx} = \frac{(1+xy)}{-x^2}

     \frac{dy}{dx} = \frac{1}{-x^2}+\frac{xy}{-x^2}

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

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

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

    now you can use the integrating factor method again as above

    where I = e^{\int\frac{1}{x}dx}

    good luck!
    Last edited by pickslides; May 26th 2009 at 03:21 PM. Reason: typo
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  5. #5
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    Quote Originally Posted by pickslides View Post
    You need to use the integrating factor method on this equation as it is not separable.

    for an equation in the from y'+f(x)y = g(x)

    your integrating factor is found as follows

    I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}

    multiplying this through the entire equation gives

    e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}

    then using the product rule, LHS changes to be

    (e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}

    now integrate both sides

    e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx

    gives y to be

    y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}

    you will have to finish the problem by using integration by parts on the numerator of the RHS.

    use f(0) = 1 to determine your constant of integration afterwards.
    Thanks so much for the help, however, just one thing that i could not understand in the algebra here:

    then using the product rule, LHS changes to be

    (e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}

    How the the LHS transform like that?
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  6. #6
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    In a sense, he is using the product rule "in reverse". The product rule states that (uv)'= u'v+ uv'.

    Here, you have e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y
    Further, it is easy to see that \left(e^{\frac{x^2}{2}}\right)'= xe^{\frac{x^2}{2}} so take u= e^{\frac{x^2}{2}} and v= y.
    Last edited by mr fantastic; June 11th 2009 at 02:17 PM. Reason: Fixed a tag
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