What is the solution for linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 ?

The solution for linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 is y=-1/13 cos(2t)-3/26 sin(2t)+1/13 e^{3t}

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linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0

linear

y^{^{''}} + 4 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

(D^{3} + D^{2} + D + 1) y = x e^{x}

30 y^{^{''}} + 25 y^{^{'}} + 4 y = 40

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

y^{^{''}} - y = sec^{3} (x) - sec (x)

y^{^{''''}} - 5 y^{^{''}} + 4 y = 8 cos (2 x) - 12 e^{- x}

What is the solution for linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 ?
The solution for linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 is y=-1/13 cos(2t)-3/26 sin(2t)+1/13 e^{3t}