What is the solution for y^{''}-2y^'-3y=24te^{2t} ?

The solution for y^{''}-2y^'-3y=24te^{2t} is y=c_{1}e^{3t}+c_{2}e^{-t}-8te^{2t}-16/3 e^{2t}

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y^{''}-2y^'-3y=24te^{2t}

y^{^{''}} - 2 y^{^{'}} - 3 y = 24 t e^{2 t}

y = c_{1} e^{3 t} + c_{2} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 24 t e^{2 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

y = c_{1} e^{3 t} + c_{2} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

y^{^{''}} - y^{^{'}} - 6 y = - 468 sin (3 t)

3 y^{^{''}} - 4 y^{^{'}} + y = 5

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- x} (9 x^{2} - 3 x - 4)

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 64 e^{t}, y (0) = 4, y^{^{'}} (0) = 7

What is the solution for y^{''}-2y^'-3y=24te^{2t} ?
The solution for y^{''}-2y^'-3y=24te^{2t} is y=c_{1}e^{3t}+c_{2}e^{-t}-8te^{2t}-16/3 e^{2t}