What is the solution for y^{''}+4y^'=6e^{3t}-3e^{-t} ?

The solution for y^{''}+4y^'=6e^{3t}-3e^{-t} is y=c_{1}+c_{2}e^{-4t}+2/7 e^{3t}+e^{-t}

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y^{''}+4y^'=6e^{3t}-3e^{-t}

y^{^{''}} + 4 y^{^{'}} = 6 e^{3 t} - 3 e^{- t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

Solution steps

y^{''} (t) + 4 y^{'} (t) = 6 e^{3 t} - 3 e^{- t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

What is the solution for y^{''}+4y^'=6e^{3t}-3e^{-t} ?
The solution for y^{''}+4y^'=6e^{3t}-3e^{-t} is y=c_{1}+c_{2}e^{-4t}+2/7 e^{3t}+e^{-t}