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Thread: Variation of Parameters Diff EQ

  1. #1
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    Variation of Parameters Diff EQ

    Hey all,

    heres the differential equation to be solved using Variation of parameter's:

    y" + y = sin(x)^2

    i got this far:

    u1'cos(2x) + u2'sin(2x) =0
    -2*u1'*sin(2x) + 2*u2'*cos(2x) = sin(x)^2

    solving for u1' gives: (-sin(x)^2*sin(2x))/2

    and u2': (-sin(x)^2*cos(2x))/2

    The final solution should take the form

    u1*cos(2x) + u2*sin(2x)

    so whats left is integrating u1' and u2'. Unfortunately due to problems with integrating and simplfying trig expressions i'm not getting the solution which is

    1/8(1-xsin(2x))

    thanks alot
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  2. #2
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    The double formulas

    This may help with your integrals

    $\displaystyle sin(2\theta)=2sin(\theta)cos(\theta)$

    and

    $\displaystyle cos(2\theta)=1-2sin^2(\theta)$

    so

    $\displaystyle u_1=\frac{1}{2}\int\sin^3(x)cos(x)dx$

    the other

    $\displaystyle u_2=-\frac{1}{2}\int\sin^2(x)(1-2sin^2(x))dx$

    You can use the reduction formula for u2 or the half angle formula

    I hope this helps
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  3. #3
    Behold, the power of SARDINES!
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    Let me see if I have this correct...

    $\displaystyle y''+y=\sin^2(x)$

    If this is the case then our complementry solution is

    $\displaystyle y=c_1\sin(x)+c_2\cos(x)$

    so computing the wronskian
    $\displaystyle \begin{vmatrix}\sin(x) & \cos(x) \\ \cos(x) & -\sin(x)\end{vmatrix}=-1$

    then

    $\displaystyle w_1=\begin{vmatrix}0 & \cos(x) \\ \sin^2(x) & -\sin(x)\end{vmatrix}=-\cos(x)\sin^2(x)$


    and

    $\displaystyle w_2=\begin{vmatrix}\sin(x) & 0 \\ cos(x)& \sin^2(x)\end{vmatrix}=\sin^3(x)$

    this would give the integrals

    $\displaystyle \int\cos(x)\sin^2(x)dx$

    and

    $\displaystyle -\int\sin^3(x)dx$
    Last edited by TheEmptySet; Feb 24th 2008 at 07:20 PM.
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  4. #4
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    i'm so sorry.

    its acutally y" +4y = sin(x) ^2

    this is why it takes the form

    c1*cos(2x) + c2(sin(2x)
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  5. #5
    Behold, the power of SARDINES!
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    Alright, lets try this again...

    $\displaystyle y''+4y=\sin^2(x)$

    If this is the case then our complementry solution is

    $\displaystyle y=c_1\sin(2x)+c_2\cos(2x)$

    so computing the wronskian
    $\displaystyle \begin{vmatrix}\sin(2x) & \cos(2x) \\ 2\cos(2x) & -2\sin(2x)\end{vmatrix}=-2$

    then

    $\displaystyle w_1=\begin{vmatrix}0 & \cos(2x) \\ \sin^2(x) & -2\sin(x)\end{vmatrix}=-\cos(2x)\sin^2(x)$


    and

    $\displaystyle w_2=\begin{vmatrix}\sin(2x) & 0 \\ 2\cos(x)& \sin^2(x)\end{vmatrix}=\sin^2(x)\sin(2x)$

    this would give the integrals

    $\displaystyle \frac{1}{2}\int\cos(2x)\sin^2(x)dx$

    and

    $\displaystyle -\frac{1}{2}\int\sin^2(x)\sin(2x)dx$

    $\displaystyle \frac{1}{2}\int\cos(2x)\sin^2(x)dx=\frac{1}{2}\int (1-2\sin^2(x))\sin^2(x)dx = $

    $\displaystyle \frac{1}{2}\int\sin^2(x)dx-\int(\sin^2(x))^2dx$

    by the double angle formula...

    $\displaystyle \frac{1}{2}\int2\sin(x)\cos(x)dx-\int(2\sin(x)\cos(x))^2dx=$

    $\displaystyle \int\sin(x)\cos(x)dx-4\int\sin^2(x)\cos^2(x)dx = $

    using the double angle formula again on sine...

    $\displaystyle \int\sin(x)\cos(x)dx-4\int2\sin(x)\cos(x)\cos^2(x)dx = $
    $\displaystyle \int\sin(x)\cos(x)dx-8\int\sin(x)\cos^3(x)dx$

    finally, now we let $\displaystyle u=\cos(x)$ and $\displaystyle du=-\sin(x)dx$
    This yeilds...$\displaystyle u_1=-\frac{1}{2}\cos^2(x)+2\cos^4(x)$

    $\displaystyle u_2$ is a bit better

    $\displaystyle -\frac{1}{2}\int\sin^2(x)\sin(2x)dx$
    rewritting with the double angle formula

    $\displaystyle -\frac{1}{2}\int\sin^2(x)(2\sin(x)\cos(x))dx=-\int\sin^3(x)\cos(x)dx$

    with $\displaystyle u=sin(x)$ and$\displaystyle du=\cos(x)dx$ we get

    $\displaystyle u_2=-\frac{1}{4}\sin^4(x)$

    I hope this gets it...
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  6. #6
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    I follow all of your work,

    but it seems when i try to get my final particular solution in the form

    yp = u1*sin(2x) + u2*cos(2x)

    i'm not getting the solution

    1/8(1-x*sin(2x))

    am i wrong in saying the particular solution yp should be in the form


    yp = u1*sin(2x) + u2*cos(2x)
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  7. #7
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    I get it now, thanks alot
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