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cke8
 one year 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
cke8
 one year 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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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1\[y''+6y'+13y=4e^{3x}\sin(2x),\; y(0)=1,\; y'(0)=2\] Homogeneous solution: \[r^2+6r+13=0\\ r=\frac{6\pm\sqrt{16}}{2}=3\pm4i\\ y_c=e^{3x}\left(C_1\cos(4x)+C_2\sin(4x)\right)\] Nonhomogeneous 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 \[y=y_c+y_p\]

cke8
 one year ago
Best ResponseYou've already chosen the best response.0Could you please explain a bit more?

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0I think the explanation provided is quite clear.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1@cke8, which part? The homogeneous solution, or the clue for the nonhomogeneous one?

cke8
 one year ago
Best ResponseYou've already chosen the best response.0The clue for the nonhomogenous one. Thanks!

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align*}y_p&=Ae^{3x}\sin(2x)+Be^{3x}\cos(2x)\\ &=e^{3x}\left(A\sin(2x)+B\cos(2x)\right)\\\\ y_p'&=3e^{3x}\left(A\sin(2x)+B\cos(2x)\right)+e^{3x}\left(2A\cos(2x)2B\sin(2x)\right)\\ &=e^{3x}\left[(3A2B)\sin(2x)+(3B+2A)\cos(2x)\right]\\\\ y_p''&=3e^{3x}\left[(3A2B)\sin(2x)+(3B+2A)\cos(2x)\right]\\&\;\;\;\;\;+e^{3x}\left[(6A4B)\cos(2x)+(6B4A)\sin(2x)\right]\\ &=e^{3x}[(5A+12B)\sin(2x)+(5B12A)\cos(2x)] \end{align*}\] 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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