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

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

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

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

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{2} e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{2} e^{2 t}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{2} e^{2 t}

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

y^{^{''}} - 3 y^{^{'}} = 8 e^{2 t} - 2 e^{- t}

y^{^{''}} - 2 y^{^{'}} + y = t

y^{^{''}} - 4 y^{^{'}} = 8 e^{6 x} (x - 1)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- x} sin (2 x)

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