What is the solution for y^{''}+2y^'+y=e^{-x}sin(x) ?

The solution for y^{''}+2y^'+y=e^{-x}sin(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-e^{-x}sin(x)

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y^{''}+2y^'+y=e^{-x}sin(x)

y^{^{''}} + 2 y^{^{'}} + y = e^{- x} sin (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} sin (x)

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = e^{- x} sin (x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} sin (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} sin (x)

\frac{d ^{2} x}{d t ^{2}} + 16 x = 1 + 2 sin (4 t)

y^{^{''}} - 4 y = 2 x - sin (x)

y^{^{''}} + y^{^{'}} - 2 y = 5

\frac{d ^{2} y}{d x ^{2}} + 9 y = x e^{2 x} + sin (3 x)

y^{^{''}} - y^{^{'}} - 12 y = 5 - 2 x

What is the solution for y^{''}+2y^'+y=e^{-x}sin(x) ?
The solution for y^{''}+2y^'+y=e^{-x}sin(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-e^{-x}sin(x)