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Math Help - Another question

  1. #1
    Member Jones's Avatar
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    Another question

    I have another question.

    y'' + 4y=2cos x + 3xsin x
    \begin{aligned}y(0) = 1\\<br />
y'(0) = 2\end{aligned}

    Show that x~sin x is a solution to this equation, and then solve the initial value problem.

    hmm, how do you go about this one?

    I thougt the solution would be on the form Ce^{-4t} + B +x~sin x
    since the solution to the homogenous equation r^2+4r=0
    \begin{aligned}<br />
r_1 = -4 \\<br />
r_2 = 0<br />
\end{aligned}
    Last edited by mr fantastic; July 23rd 2009 at 06:41 PM. Reason: Moved from original thread
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jones View Post
    I have another question.

    y'' + 4y=2cos x + 3xsin x
    \begin{aligned}y(0) = 1\\<br />
y'(0) = 2\end{aligned}

    Show that x~sin x is a solution to this equation, and then solve the initial value problem.

    hmm, how do you go about this one?

    I thougt the solution would be on the form Ce^{-4t} + B +x~sin x
    since the solution to the homogenous equation r^2+4r=0
    \begin{aligned}<br />
r_1 = -4 \\<br />
r_2 = 0<br />
\end{aligned}
    Be careful! The auxiliary equation is r^2+4=0\implies r=\pm2i

    So the complimentary solution will have the form y_c=c_1\cos\left(2x\right)+c_2\sin\left(2x\right).

    Now, (via annihilator approach as described here - try to fill in what's missing), the particular solution is of the form y_p=A\cos x+B\sin x+Cx\cos x+ Dx\sin x

    So, it follows that

    y_p^{\prime}=-A\sin x+B\cos x+C\left(\cos x-x\sin x\right)+D\left(\sin x+x\cos x\right) =\left(D-A\right)\sin x+\left(B+C\right)\cos x-Cx\sin x+Dx\cos x

    y_p^{\prime\prime}=-A\cos x-B\sin x+C\left(-2\sin x-x\cos x\right)+D\left(2\cos x-x\sin x\right) =\left(-B-2C\right)\sin x+\left(2D-A\right)\cos x-Cx\cos x-Dx\sin x

    Thus, y_p^{\prime\prime}+4y_p=2\cos x+3x\sin x \implies \left(3B-2C\right)\sin x+\left(3A+2D\right)\cos x+3Cx\cos x+3Dx\sin x=2\cos x+3x\sin x

    Comparing coefficients, we have

    \begin{aligned}<br />
   3B-2C & = 0\\<br />
   3A+2D & = 2\\<br />
   3C & = 0\\<br />
   3D & = 3<br />
\end{aligned}

    From this, it follows that A=B=C=0,D=1

    So the solution will have the form y(x)=c_1\cos\left(2x\right)+c_2\sin\left(2x\right)  +x\sin x.

    Now apply the initial conditions to find c_1 and c_2

    Can you take it from here?
    Last edited by mr fantastic; July 23rd 2009 at 06:43 PM. Reason: Edited quote
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  3. #3
    Member Jones's Avatar
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    Thanks a lot!
    Yes i think i can manage from here.

    Ugh, gotta be careful about those pitfalls. =/
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    I have another question.

    y'' + 4y=2cos x + 3xsin x
    \begin{aligned}y(0) = 1\\<br />
y'(0) = 2\end{aligned}

    Show that x~sin x is a solution to this equation, and then solve the initial value problem.

    hmm, how do you go about this one?

    I thougt the solution would be on the form Ce^{-4t} + B +x~sin x
    since the solution to the homogenous equation r^2+4r=0
    \begin{aligned}<br />
r_1 = -4 \\<br />
r_2 = 0<br />
\end{aligned}
    When a question is expressed like this you generally just show that the suggested solution is in fact a solution.

    So put:

    y(x)=x\;\sin(x)

    Then;

     <br />
y'=\sin(x)+x\cos(x)<br />

     <br />
y''=2\cos(x)-x\sin(x)<br />

    Then:

     <br />
y''+4y=2\cos+3x\sin(x)<br />

    As required.

    What this shows is that y(x)=x\;\sin(x) is a particular integral for this ODE. To find the solution to the IVP you now need only solve the homogeneous ODE add its general solution to the PI and apply the initial conditions.

    CB
    Last edited by mr fantastic; July 23rd 2009 at 06:44 PM. Reason: Edited quote
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