What is the solution for 2y^{''}+2y=-16cos(3x) ?

The solution for 2y^{''}+2y=-16cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+cos(3x)

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2y^{''}+2y=-16cos(3x)

2 y^{^{''}} + 2 y = - 16 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + cos (3 x)

Solution steps

2 y^{''} (x) + 2 y = - 16 cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + cos (3 x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 2

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x^{2} + 6 e^{x}

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

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

What is the solution for 2y^{''}+2y=-16cos(3x) ?
The solution for 2y^{''}+2y=-16cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+cos(3x)