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

The solution for y^{''}+4y=e^{3t}+4cos(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/13 e^{3t}-4/5 cos(3t)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{13} e^{3 t} - \frac{4}{5} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{13} e^{3 t} - \frac{4}{5} cos (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{13} e^{3 t} - \frac{4}{5} cos (3 t)

y^{^{''}} (s) + y (s) = \frac{1}{s ^{2}} - \frac{2}{s ^{2}}

y^{^{''}} - y^{^{'}} - 2 y = - 2 x^{3} - 3^{2} + 8 x + 1

y^{^{''}} - 36 y = 9 x e^{6 x}

31 y^{^{''}} + 800 y = 100 cos (4 t), y (0) = 0, y^{^{'}} (0) = 0

x^{^{''''}} + 2 x^{^{''}} + x = 4 t e^{t}

What is the solution for y^{''}+4y=e^{3t}+4cos(3t) ?
The solution for y^{''}+4y=e^{3t}+4cos(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/13 e^{3t}-4/5 cos(3t)