Solve both ways: a) y" -2y' + y = e^2x b) Solve only by variation of parameters

Solve both ways:

a) y" -2y' + y = e^2x

b) Solve only by variation of parameters

b) y" -9y = x/(e^3x)

c) y" -2y' + y = (e^x)/(x^4)

d) y" + y = sec^3 x

Expert Solution

Colby Messinger answered 2 years ago

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