What is the solution for y^{''}+2y^'+y=4sin(x) ?

The solution for y^{''}+2y^'+y=4sin(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-2cos(x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y - 24 = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 t

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 24 e^{3 x}

y^{^{''}} - y^{^{'}} - 2 y = 8 sin (t) - 2 cos (t)

y^{^{''}} + 6 y^{^{'}} + 13 y = 13 x^{2}

What is the solution for y^{''}+2y^'+y=4sin(x) ?
The solution for y^{''}+2y^'+y=4sin(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-2cos(x)