What is the solution for y^{''}-3y^'=8e^{2t}-2e^{-t} ?

The solution for y^{''}-3y^'=8e^{2t}-2e^{-t} is y=c_{1}e^{3t}+c_{2}-4e^{2t}-1/2 e^{-t}

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

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

y = c_{1} e^{3 t} + c_{2} - 4 e^{2 t} - \frac{1}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} - 4 e^{2 t} - \frac{1}{2} e^{- t}

y = c_{1} e^{3 t} + c_{2} - 4 e^{2 t} - \frac{1}{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)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 8 e^{x}

\frac{d ^{2} y}{d t ^{2}} = 0.5 t - y

What is the solution for y^{''}-3y^'=8e^{2t}-2e^{-t} ?
The solution for y^{''}-3y^'=8e^{2t}-2e^{-t} is y=c_{1}e^{3t}+c_{2}-4e^{2t}-1/2 e^{-t}