Results 1 to 7 of 7

Math Help - Variation of Parameters Diff EQ

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    16

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    The double formulas

    This may help with your integrals

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

    and

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

    so

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

    the other

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Let me see if I have this correct...

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

    If this is the case then our complementry solution is

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

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

    then

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


    and

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

    this would give the integrals

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

    and

    -\int\sin^3(x)dx
    Last edited by TheEmptySet; February 24th 2008 at 07:20 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2008
    Posts
    16
    i'm so sorry.

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

    this is why it takes the form

    c1*cos(2x) + c2(sin(2x)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Alright, lets try this again...

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

    If this is the case then our complementry solution is

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

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

    then

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


    and

    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

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

    and

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

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

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

    by the double angle formula...

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

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

    using the double angle formula again on sine...

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

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

    u_2 is a bit better

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

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

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

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

    I hope this gets it...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2008
    Posts
    16
    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)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2008
    Posts
    16
    I get it now, thanks alot
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 13th 2010, 04:18 PM
  2. Variation of Parameters
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 26th 2010, 09:21 AM
  3. Replies: 0
    Last Post: May 29th 2008, 11:42 AM
  4. Variation of parameters Diff EQ
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 29th 2008, 05:11 AM
  5. Variation of Parameters
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 4th 2007, 08:44 PM

Search Tags


/mathhelpforum @mathhelpforum