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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
 one year ago
 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
 one year ago
 one year ago

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SithsAndGigglesBest 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 \[y=y_c+y_p\]
 one year ago

cke8Best ResponseYou've already chosen the best response.0
Could you please explain a bit more?
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
I think the explanation provided is quite clear.
 one year ago

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

cke8Best ResponseYou've already chosen the best response.0
The clue for the nonhomogenous one. Thanks!
 one year ago

SithsAndGigglesBest 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.
 one year ago
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