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

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

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{4} e^{2 t} - \frac{4}{5} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{4} e^{2 t} - \frac{4}{5} cos (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{4} e^{2 t} - \frac{4}{5} cos (3 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- x} (3 - 4 x)

y^{^{''}} - 16 y^{^{'}} = 2 e^{4 x}

y^{^{'''}} + 4 y^{^{'}} = t^{2} + 2 t

y^{^{''}} - y = sin (t), y (0) = 1, y^{^{'}} (0) = 1

4 y^{^{''}} + 8 y^{^{'}} + 5 y = e^{- x} sec (\frac{x}{2})

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