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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
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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SithsAndGiggles
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Could you please explain a bit more?

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

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

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

SithsAndGiggles
 2 years 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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