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

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

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y^{''}+4y=3e^{3t}

y^{^{''}} + 4 y = 3 e^{3 t}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 18 e^{2 x} - 4 sin (x) - x^{2}

3 y^{^{''}} - 2 y - y = 2 t^{2} - t - 3 e^{- 3 t}

y^{^{''}} - 2 y^{^{'}} - 15 y = (6 t + 3 tt + 1) e^{t}

25 y + 12 y^{^{'}} + 4 y^{^{''}} = - 2

y^{^{'''}} - y^{^{'}} = 4 x^{2} + 5 x - 2

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