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Thread: Finding the PS (of a DE)

  1. #1
    Nov 2008

    Finding the PS (of a DE)

    Use a suitable trial function to find the particular solution of

    $\displaystyle \frac{dy}{dx}+6y=3sin(4x)$

    $\displaystyle y=Acos(4x)+Bsin(4x)$

    $\displaystyle \frac{dy}{dx}=-4Asin(2x)+4Bcos(2x)$

    From the equation, $\displaystyle -4Asin(2x)+4Bcos(2x)+6[Acos(2x)+Bsin(2x)]=3sin(4x)$

    Now I don't understand the last bit where you have to pick the two parts which correspond to 0 and 3. Could someone explain it for me?
    Last edited by Haris; Feb 11th 2009 at 12:28 PM.
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  2. #2
    Feb 2009
    For an equation of the form:
    $\displaystyle \mathsf{\frac{dy}{dx}+P(x)\cdot y = Q(x)}$ (in this case $\displaystyle \mathsf{P(x)=6}$ and $\displaystyle \mathsf{Q(x)=5\sin{2x}}$)

    $\displaystyle \mathsf{y=\frac{\int Q(x)\cdot e^{\int P(x)dx}dx}{e^{\int P(x)dx}}}$

    For this case:

    $\displaystyle \mathsf{\int P(x)dx=6x}$ and hence $\displaystyle \mathsf{e^{\int P(x)dx}=e^{6x}}$

    So to the guts of the problem (remembering the constant of integration at this stage):

    $\displaystyle \mathsf{y=\frac{\int 5\sin{2x}\cdot e^{6x}dx}{e^{6x}}=\frac{e^{6x}\cdot \frac{3}{4}\sin{2x}-e^{6x}\cdot \frac{1}{4}\cos{2x}+C}{e^{6x}}=\frac{3}{4}\sin{2x}-\frac{1}{4}\cos{2x}+Ce^{-6x}}$

    $\displaystyle \mathsf{\frac{dy}{dx}=\frac{3}{2}\cos{2x}+\frac{1} {2}\sin{2x}-6Ce^{-6x}}$


    $\displaystyle \mathsf{\frac{dy}{dx}+6y=5\sin{2x}}$

    $\displaystyle \mathsf{\frac{3}{2}\cos{2x}+\frac{1}{2}\sin{2x}-6Ce^{-6x}+6(\frac{3}{4}\sin{2x}-\frac{1}{4}\cos{2x}+Ce^{-6x})=5\sin{2x}}$

    To determine C you'll need a known point
    Last edited by danothy; Feb 10th 2009 at 04:05 PM. Reason: Typo noticed in original post
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