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}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} 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} t

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

y^{^{''}} - 2 y - 4 x^{2} e^{2 x} = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 3, y^{^{'}} (0) = 1

y^{^{''}} - 4 y = 12 e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 4 (x - 1) \cdot e^{- x} - 1

2 y^{^{''}} + 6 y^{^{'}} = 5

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}t