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)

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} = 40 x - 7 - 6 e^{2 x}

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{9 x}) + 9 (cos (x))

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 20 cos (x)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 50

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)