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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{x ^{4} + 2 x ^{2} - 1}{x} d x

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{x ^{3} + x - 2})

n = 21 \sum \infty e^{- 0.06 n}

(x^{2} - 9) \frac{d y}{d x} + x y = 0

\int 16 - 4 x^{2} d x

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