HELP ME

Dingus - March 13, 2025

I am stuck on this question. It is dy/dx ^ 2 - 2dy/dx + 5y = e^x * sinx. I know that I have to find the roots first so it is m^2 - 2m + 5 but how do I find the annihilator for e^x * sinx????? help

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ryan0 - March 14, 2025, 6:33 a.m.

the annihilator for e^x sinx is (D-1)^2 + 1. The formula I try to remember when I come across lets say e^ax * sinbx is (D-a)^2 + b^2!

Dingus - March 15, 2025, 9:05 p.m.

thanks :O