What is the solution for y^{''}+9y=cos(t) ?

The solution for y^{''}+9y=cos(t) is y=c_{1}cos(3t)+c_{2}sin(3t)+1/8 cos(t)

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y^{''}+9y=cos(t)

y^{^{''}} + 9 y = cos (t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} cos (t)

Solution steps

y^{''} (t) + 9 y = cos (t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} cos (t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} cos (t)

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

y^{^{''}} + 4 y = t^{3} + 3 e^{t}, y (0) = 0, y^{^{'}} (0) = 2

4 y^{^{''}} - y = 12 x + 6 x

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 e^{- 4 t}

1 x^{^{''}} + 2 x = 8 sin (2 t)

What is the solution for y^{''}+9y=cos(t) ?
The solution for y^{''}+9y=cos(t) is y=c_{1}cos(3t)+c_{2}sin(3t)+1/8 cos(t)