What is the solution for y^{''}+y+2sin(x)=0 ?

The solution for y^{''}+y+2sin(x)=0 is y=c_{1}cos(x)+c_{2}sin(x)+xcos(x)

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y^{''}+y+2sin(x)=0

y^{^{''}} + y + 2 sin (x) = 0

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

Solution steps

y^{''} (x) + y + 2 sin (x) = 0

Solve linear ODE:

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

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

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

(D^{2} + 1) y = e^{- x} + cos (x) + x^{3}

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

y^{^{''}} + y^{^{'}} = x (e^{3 x} + 1)

f^{^{''}} (t) + 9 \cdot f (t) = (- 2 \cdot e^{(3 t)})

What is the solution for y^{''}+y+2sin(x)=0 ?
The solution for y^{''}+y+2sin(x)=0 is y=c_{1}cos(x)+c_{2}sin(x)+xcos(x)