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

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

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y^{''}+2y^'+y=cos(x)

y^{^{''}} + 2 y^{^{'}} + y = cos (x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = cos (x)

Solve linear ODE:

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

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

y^{^{''}} + 4 y = g (t), y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 6 y = 3 x^{2} - 8

y^{^{'''}} + 8 y = 2 x - 5 + 8 e^{- 2} x

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = x e^{x}

y^{^{''}} - 4 y = 4 e^{3 x}

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