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

The solution for y^{''}+2y^'=-sin(x) is y=c_{1}+c_{2}e^{-2x}+1/5 sin(x)+2/5 cos(x)

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 + 2 x + 2 x^{2}

y^{^{''}} - 9 y = sin (2 x)

\frac{d ^{2} y}{d x ^{2}} + y = 32 x^{3}

y^{^{''}} - 0.16 y = 32 cosh (0.4 x)

y^{^{''}} + 9 y = e^{4 x} + 3 x^{2} - 5

What is the solution for y^{''}+2y^'=-sin(x) ?
The solution for y^{''}+2y^'=-sin(x) is y=c_{1}+c_{2}e^{-2x}+1/5 sin(x)+2/5 cos(x)