Solve the differential equation: \newline $y'' + 8y' + 41y = 3 + e^{-x}$
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The characteristic equation is r^2 + 8r + 4 = 0. Using the quadratic formula, we find that the roots are r = (-8 ± √(8^2 - 4*4))/2 = -4 ± √6. Therefore, the complementary solution is y_c = c1e^((-4 + √6)x) + c2e^((-4 - √6)x), where c1 and c2 are constants. Show more…
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