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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 10 x^{2} - 2 x e^{2}

y^{^{''}} - y^{^{'}} - 6 y = 22 cos (3 x)

9 y^{^{''}} + 6 y^{^{'}} + y = 2, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = t

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