What is the solution for y^{''}+4y=cos(3t) ?

The solution for y^{''}+4y=cos(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/5 cos(3t)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

What is the solution for y^{''}+4y=cos(3t) ?
The solution for y^{''}+4y=cos(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/5 cos(3t)