What is the solution for y^{''}+3y=-48t^2e^3t ?

The solution for y^{''}+3y=-48t^2e^3t is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)-16e^3t^3+32e^3t

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y^{''}+3y=-48t^2e^3t

y^{^{''}} + 3 y = - 48 t^{2} e^{3} t

y = c_{1} cos (3 t) + c_{2} sin (3 t) - 16 e^{3} t^{3} + 32 e^{3} t

Solution steps

y^{''} (t) + 3 y = - 48 t^{2} e^{3} t

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) - 16 e^{3} t^{3} + 32 e^{3} t

y = c_{1} cos (3 t) + c_{2} sin (3 t) - 16 e^{3} t^{3} + 32 e^{3} t

y^{^{''''}} - 256 y = 2 e^{- 4 x} + 3 e^{3 x} + cos (4 x) + 2

y^{^{''}} + y = 3 e^{- 6 x}, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3} t

y^{^{''}} - 10 y^{^{'}} + 25 y = 132 x^{2} (e^{(5 x)})

What is the solution for y^{''}+3y=-48t^2e^3t ?
The solution for y^{''}+3y=-48t^2e^3t is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)-16e^3t^3+32e^3t