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cke8

  • 2 years ago

Please help me solve y''(x)+6 y'(x)+13 y(x) = 4 e^(-3 x) sin(2 x), y(0) = 1, y'(0) = 2

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  1. SithsAndGiggles
    • 2 years ago
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    \[y''+6y'+13y=4e^{-3x}\sin(2x),\; y(0)=1,\; y'(0)=2\] Homogeneous solution: \[r^2+6r+13=0\\ r=\frac{-6\pm\sqrt{-16}}{2}=-3\pm4i\\ y_c=e^{-3x}\left(C_1\cos(4x)+C_2\sin(4x)\right)\] Non-homogeneous solution: Using the method of undetermined coefficients, try (as a guess) \[y_p=Ae^{-3x}\sin(2x)+Be^{-3x}\cos(2x)\] I'll leave that work to you, since it'd be a pain to type it all out, but I'll be available to check what you get later. The general solution will then be \[y=y_c+y_p\]

  2. cke8
    • 2 years ago
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    Could you please explain a bit more?

  3. abb0t
    • 2 years ago
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    I think the explanation provided is quite clear.

  4. SithsAndGiggles
    • 2 years ago
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    @cke8, which part? The homogeneous solution, or the clue for the non-homogeneous one?

  5. cke8
    • 2 years ago
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    The clue for the non-homogenous one. Thanks!

  6. SithsAndGiggles
    • 2 years ago
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    \[\begin{align*}y_p&=Ae^{-3x}\sin(2x)+Be^{-3x}\cos(2x)\\ &=e^{-3x}\left(A\sin(2x)+B\cos(2x)\right)\\\\ y_p'&=-3e^{-3x}\left(A\sin(2x)+B\cos(2x)\right)+e^{-3x}\left(2A\cos(2x)-2B\sin(2x)\right)\\ &=e^{-3x}\left[(-3A-2B)\sin(2x)+(-3B+2A)\cos(2x)\right]\\\\ y_p''&=-3e^{-3x}\left[(-3A-2B)\sin(2x)+(-3B+2A)\cos(2x)\right]\\&\;\;\;\;\;+e^{-3x}\left[(-6A-4B)\cos(2x)+(6B-4A)\sin(2x)\right]\\ &=e^{-3x}[(5A+12B)\sin(2x)+(5B-12A)\cos(2x)] \end{align*}\] Now plug in y_p'', y_p', and y_p into the original equation and solve for A and B. It seems daunting, but doable. Be sure to keep track of your coefficients and their signs.

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