Find the general solution: y + 4y' + 5y = xe^(-2x) cos(x)
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The characteristic equation is r^2 + 4r + 5 = 0. Using the quadratic formula, we find that the roots are r = (-4 ± √(4^2 - 4(1)(5))) / (2(1)) = (-4 ± √(-4)) / 2 = -2 ± i. Therefore, the complementary solution is y_c = c1e^(-2x)cos(x) + c2e^(-2x)sin(x), where Show more…
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