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

The solution for y^{''}+2y=sin(x) is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)+sin(x)

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + sin (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + sin (x)

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

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{x} cos (2 x)

y^{^{''}} + 2 y^{^{'}} + y = e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

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

y^{^{'''}} - 25 y^{^{'}} = t e^{- 5 t} + 6 cos (5 t)

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