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

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

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + x - 4 - \frac{1}{2} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + x - 4 - \frac{1}{2} cos (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} + x - 4 - \frac{1}{2} cos (x)

y^{^{''}} + 7 y^{^{'}} - 12 = e^{3 x} - 2 e^{x} - 5 x

y^{^{''}} + 2 y^{^{'}} + 2 y = 1 + cos (2 t)

y^{^{''}} + 4 y^{^{'}} + 7 t = t^{3} e^{4 t} cos (t)

\frac{d ^{2} y}{d x ^{2}} - y = sin^{2} (x)

y^{^{''}} - 2 y^{^{'}} + 26 y = e x cos (5 x)

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