What is the solution for y^{''}-2y^'-3y=4e^t ?

The solution for y^{''}-2y^'-3y=4e^t is y=c_{1}e^{3t}+c_{2}e^{-t}-e^t

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y^{''}-2y^'-3y=4e^t

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 4 e^{t}

Solve linear ODE:

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

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

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

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = 2 e^{- x}

y^{^{''}} - 3 y^{^{'}} - 28 y = 528 e^{- t}

y^{^{''}} + y^{^{'}} + y = e^{3 x} + 6 e^{x} - 3 e^{- 2 x} + 5

d (x) = 5 (x - 3)^{2} + 1

What is the solution for y^{''}-2y^'-3y=4e^t ?
The solution for y^{''}-2y^'-3y=4e^t is y=c_{1}e^{3t}+c_{2}e^{-t}-e^t