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Thread: Can't get this Differential Eq

  1. #1
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    Unhappy Can't get this Differential Eq

    Find a particular solution for y"+y=sinx+xcosx

    I used an initial guess of

    yp(x)= Asinx+Bcosx+Cxcosx+Dxsinx

    y'p(x)= Acosx-Bsinx-Cxsinx+Ccosx+Dxcosx+Dsinx

    y"p(x)= -Asinx-Bcosx-Cxcosx-Csinx-Csinx-Dxsinx+Dcosx+Dcosx

    I end up with the following system of equations to solve my coefficients

    -A+A-C-C (sinx)=1(sinx)

    -B+D+D+B(cosx)=0(cosx)

    -C+C=1 (xcosx) How can this be??? 0=1??

    -D+D=0 (xsinx)

    Here's where I hit my snag. The answer has an x^2 power in it too. I'm stumped!
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  2. #2
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    Hello, doke!

    Find a particular solution for: .$\displaystyle y'' + y\:=\:\sin x + x\cos x$

    We already know that: $\displaystyle y \:=\:C_1\cos x + C_2\sin x$ is a homogenous solution.


    Due to the $\displaystyle y''$, we must begin "two level higher".

    We will try:
    . . $\displaystyle y \:=\:Ax\sin x + Bx\cos x + Cx^2\sin x + Dx^2\cos x$

    Then:
    . . $\displaystyle y'\:=\:A\sin x + B\cos x + (C-2B)\sin x + (A + 2D)x\cos x - Dx^2\sin x + Cx^2\cos x$

    And:
    . . $\displaystyle y'' \:=\:2(C - B)\sin x + 2(A + D)\cos x - (A + 4D)\sin x \,+$ $\displaystyle (4C - B)x\cos x - Cx^2\sin x + Dx^2\cos x$


    Hence:
    . . $\displaystyle y'' + y \;=\;2(C - B)\sin x + 2(A + D)\cos x - 4Dx\sin x + 4Cx\cos x$ $\displaystyle =\;\sin x + x\cos x$

    Then: .$\displaystyle \begin{Bmatrix}2(C - B) = 1 \\2(A + D) = 0\\ -4D = 0 \\ 4C = 1\end{Bmatrix}$

    Solve the system and get: .$\displaystyle A = 0,\;\;B = \text{-}\frac{1}{4},\;\;C = \frac{1}{4},\;\;D = 0$

    Therefore, the particular solution is: .$\displaystyle \boxed{y_p \:=\:-\frac{1}{4}x\cos x + \frac{1}{4}x^2\sin x}$

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  3. #3
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    Wow! Thank you very much. I do have a question though... as a matter of general procedure, I suppose I could have used Asinx + Bcosx +Cxsinx +Dxcosx + Ex^2sinx + Fx^2cosx correct? I guess the extra terms will probably drop out but it seems more "consistent" to me.

    I've done other problems such as y" -4y = sinhx and didn't start "two levels higher". Why did those work out okay and this one didn't?

    Thank you in advance.
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  4. #4
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    Quote Originally Posted by doke View Post
    I've done other problems such as y" -4y = sinhx and didn't start "two levels higher". Why did those work out okay and this one didn't?
    The equation is,
    $\displaystyle y''-4y=\frac{1}{2}e^x-\frac{1}{2}e^{-x}$
    The charachteristic equation is,
    $\displaystyle k^2-4=0$
    The solutions are,
    $\displaystyle k=\pm 2$
    Thus, the two linearly independant solutions are,
    $\displaystyle e^{2x} \mbox{ and }e^{-2x}$
    That means when you search for a particular solution to the non-homogenous equation you look for,
    $\displaystyle Ae^{x}+Be^{-x}$
    Since, $\displaystyle e^x,e^{-x}$ are not part of the two linearlt independant solutions.
    However you had, $\displaystyle \sinh (2x)$ then you need to "raise one level" because, $\displaystyle e^{2x}, e^{-2x}$ are the independant solutions.
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