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Math Help - Help with differential equations

  1. #1
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    Help with differential equations

    1. (x+a) y' -3y = (x+a)^4

    2. Interchange dependent and independent variables to solve:

    (2x-5y^3) y' + y = 0, y(1)=1

    3. Using the subst. z=y^-2 solve:

    y' = xy - y^3 e^-x^2

    4. Using the subst. 1+x = e^t solve:

    (1+x)^2 y" + (1+x) y' + y =4cos(ln(1+x))

    Any help will be much appreciated.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Happy Dancer View Post
    1. (x+a) y' -3y = (x+a)^4

    Any help will be much appreciated.
    Divide the equation through by (x+a).

    We should then get \frac{\,dy}{\,dx}-\frac{3}{x+a}y=(x+a)^3

    Apply the integrating factor technique, where P(x)=-\frac{3}{x+a}.

    Can you take it from here?

    --Chris
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Happy Dancer View Post
    3. Using the subst. z=y^-2 solve:

    y' = xy - y^3 e^-x^2
    What you need to do here is rewrite the equation so the DE is in terms of z and x.

    Since z=y^{-2}\implies y=z^{-\frac{1}{2}}, we need to find \frac{\,dy}{\,dx}. Applying the chain rule, we know that \frac{\,dy}{\,dx}=\frac{\,dy}{\,dz}\cdot\frac{\,dz  }{\,dx}=-\tfrac{1}{2}z^{-\frac{3}{2}}\frac{\,dz}{\,dx}

    Now make all these substitutions into the equation:

    -\tfrac{1}{2}z^{-\frac{3}{2}}\frac{\,dz}{\,dx}=xz^{-\frac{1}{2}}-z^{-\frac{3}{2}}e^{-x^2}.

    Multiplying both sides by -2z^{\frac{3}{2}}, we get

    \frac{\,dz}{\,dx}=-2xz+2e^{-x^2}\implies \frac{\,dz}{\,dx}+2xz=2e^{-x^2}

    This is a linear equation. Apply the integrating factor, where P(x)=2x.

    Can you take it from here?

    --Chris
    Last edited by Chris L T521; August 15th 2008 at 04:57 PM.
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  4. #4
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    Thanks, I got the answers to be y=x(x+a)^3 + C(x+a)^3, y=1/(SQRT(2(x+C)e^-x^2)) and y=A(1+x)^i + B/(1+x)^i +4sin(ln(1+x)), for 1, 3, 4 respectively. Could someone be kind enough to check these answers for me please. But I'm still having problems with question 2, if someone could please help that would be much appreciated.
    Last edited by Happy Dancer; August 17th 2008 at 03:42 PM.
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  5. #5
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    Quote Originally Posted by Happy Dancer View Post
    Thanks, I got the answers to be y=x(x+a)^3 + C(x+a)^3, y=1/(SQRT(2(x+C)e^-x^2)) and y=A(1+x)^i + B/(1+x)^i +4sin(ln(1+x)), for 1, 2, 4 respectively. Could someone be kind enough to check these answers for me please. But I'm still having problems with question 2, if someone could please help that would be much appreciated.
    You can check your answers by carefully substituting them into the DE and seeing whether or not you get equality.

    By the way, what's with the exponents I've highlighted in red ...... ?
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  6. #6
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    I got them to be imaginary/complex
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  7. #7
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    Can anyone help me to interchange the dependent and independent variables to solve Q2 though. Thanks
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  8. #8
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    Quote Originally Posted by Happy Dancer View Post
    [snip]

    4. Using the subst. 1+x = e^t solve:



    (1+x)^2 y" + (1+x) y' + y =4cos(ln(1+x))



    [snip]


    Quote Originally Posted by Happy Dancer View Post
    Thanks, I got the answers to be [snip] y=A(1+x)^i + B/(1+x)^i +4sin(ln(1+x)), for [snip] 4 [snip]
    Quote Originally Posted by Mr Fantastic
    By the way, what's with the exponents I've highlighted in red ...... ?
    Quote Originally Posted by Happy Dancer View Post
    I got them to be imaginary/complex
    The solution to the DE does not contain non-real exponents.

    I suggest you review the general solution to a second order DE with constant coefficients, particularly the bit about the solutions to the auxillary equation being non-real.

    The homogeneous solution to \frac{d^2 y}{dt^2} + y = 4 \cos t is y = A \cos t + B \sin t.
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  9. #9
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    Ok thanks forgot about Euler's theorem. So I got the answer to be

    y = (C+8)cos(ln(1+x)) + (D+4)sin(ln)1+x))

    But I'm still not 100% sure about what to do when PI is part of CF.

    Also would you be kind enough to start me off in the right direction for Q2.
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