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Math Help - [SOLVED] variation of parameters

  1. #1
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    [SOLVED] variation of parameters

    Could someone show me where I am going wrong? I need to solve this using varation of parameters.

    y''+y=cos(x)

    First I found the solution to the homogeneous equation:
    y''+y=0
    m^{2}+1=0
    y_{c}=c_{1}sin(x)+c_{2}cos(x)

    Next, (note that y' is hard to see in math form)**
    y_{p}=u_{1}cos(x)+u_{2}sin(x)
    y_{p}'=u_{1}'cos(x)-u_{1}sin(x)+u_{2}'sin(x)+u_{2}cos(x)

    Next I set this to 0:
    u_{1}'cos(x)+u_{2}'sin(x)=0

    This changes y_{p} to:
    y_{p}'=-u_{1}sin(x)+u_{2}cos(x)

    y''_{p}=-u_{1}'sin(x)-u_{1}cos(x)+u_{2}'cos(x)-u_{2}sin(x)

    Next I plugged y_{p},y'_{p},y''_{p} into the Origioal DE for  y,y'',y'' respectively.

    My result is:
    -u_{1}'sin(x)+u'_{2}cos(x)=cos(x)
    Second equation from before:
    u'_{1}cos(x)+u'_{2}sin(x)=0

    Solving these 2 equation for u'_{1} and  u'_{2} I get:
    u'_{1}=-sin(x)cos(x)
     u'_{2}=cos^{2}(x)

    Integrating:
    u_{1}=\dfrac{cos^{2}(x)}{2}
    u_{2}=\dfrac{x}{2}+\dfrac{1}{2}sin(x)cos(x)

    Using the formula:
    y=y_{c}+y_{p}
    y=c_{1}sin(x)+c_{2}cos(x)+\dfrac{cos^{2}(x)}{2}+\d  frac{x}{2}+\dfrac{1}{2}sin(x)cos(x)

    The solution should be: y=c_{1}sin(x)+c_{2}cos(x) +\dfrac{1}{2}x sin(x)
    This is according to wolfram alpha:http://www.wolframalpha.com/input/?i...y%3Dcos%28x%29
    Last edited by snaes; March 16th 2010 at 03:32 PM. Reason: note**
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  2. #2
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    You almost made it. Your solution for u_1\text{ and }u_2 is correct, but you need to plug those into the equation for y_p:

    y_p=u_1\cos(x)+u_2\sin(x)

    y_p=\frac{cos^{2}(x)}{2}cos(x)+\left(\frac{x}{2}+\  frac{1}{2}cos(x)sin(x)\right)sin(x)

    y_p=\frac{cos^{2}(x)}{2}cos(x)+\frac{x}{2}sin(x)+\  frac{1}{2}cos(x)sin^{2}(x)

    and the first and third terms combine using cos^2(x)+sin^2(x)=1

    y_p=\frac{1}{2}cos(x)+\frac{1}{2}xsin(x)

    And the first term is absorbed into the homogeneous solution.

    Post again if you are still having trouble.
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  3. #3
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    thanks!

    Now i know how to do this. Thanks.

    There was one other part of the problem that I was a little unsure of:
    First I found the solution to the homogeneous equation:




    Okay, this should be simple but I am not quite sure.
    m=\sqrt{i} or m=-\sqrt{i}

    That makes the solution to the homogeneous equation:
    y_{c}=c_{1}e^{ix}+c_{2}e^{-ix}

    Using euler's method:
    y_{c}=c_{1}(cos(x)+i sin(x))+c_{2}(cos(-x)+i sin(-x))
    y_{c}=c_{1}cos(x)+c_{1}isin(x)+c_{2}cos(x)-c_{2}isin(x)

    How does this simplify to ?
    Or mabye this process is wrong..
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  4. #4
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    Quote Originally Posted by snaes View Post
    Now i know how to do this. Thanks.

    There was one other part of the problem that I was a little unsure of:
    First I found the solution to the homogeneous equation:




    Okay, this should be simple but I am not quite sure.
    m=\sqrt{i} or m=-\sqrt{i}

    That makes the solution to the homogeneous equation:
    y_{c}=c_{1}e^{ix}+c_{2}e^{-ix}

    Using euler's method:
    y_{c}=c_{1}(cos(x)+i sin(x))+c_{2}(cos(-x)+i sin(-x))
    y_{c}=c_{1}cos(x)+c_{1}isin(x)+c_{2}cos(x)-c_{2}isin(x)

    How does this simplify to ?
    Or mabye this process is wrong..
    you are almost there


    y_{c}=c_{1}cos(x)+c_{1}isin(x)+c_{2}cos(x)-c_{2}isin(x)

    y = (c_{1} + c_{2})\cos{x} + (c_{1}i - c_{2}i)\sin{x}

    now you can just say

    c_{1} + c_{2} = c_{3}

    and

    c_{1}i - c_{2}i = c_{4}

    combine that all together and you get

    y= c_{3}\cos{x} + c_{4}\sin{x}
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  5. #5
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    Thanks!

    Thanks for finishing that off for me.
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