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

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

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y^{''}+4y=-cos(3t)-2sin(3t)

y^{^{''}} + 4 y = - cos (3 t) - 2 sin (3 t)

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

Solution steps

y^{''} (t) + 4 y = - cos (3 t) - 2 sin (3 t)

Solve linear ODE:

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

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

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