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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - y = e^{\frac{x}{2}} + 7

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} cos (2 x)

y^{^{''}} + y^{^{'}} - 6 y = 2 x^{2} + 1

y^{^{''}} + 4 y = 3 sin (3 t)

y^{^{''}} + 9 y = cos (x) + \frac{1}{3} cos (3 x)

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