4) Determine the general solution of $y'' - 2y' + 2y = an(x)e^x$
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y'' - 2y' + 2y = 0 r^2 - 2r + 2 = 0 Using the quadratic formula, we get r = (2 ± sqrt(4 - 8i))/2 r = 1 ± sqrt(1 - 2i) Since the discriminant is negative, we have complex roots. Let's write them in polar form: r1 = 1 + sqrt(2)i = 2e^(iπ/4) r2 = 1 - Show more…
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