What is the solution for y^{''}-5y^'=8e^{4t}-4e^{-t} ?

The solution for y^{''}-5y^'=8e^{4t}-4e^{-t} is y=c_{1}e^{5t}+c_{2}-2e^{4t}-2/3 e^{-t}

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y^{''}-5y^'=8e^{4t}-4e^{-t}

y^{^{''}} - 5 y^{^{'}} = 8 e^{4 t} - 4 e^{- t}

y = c_{1} e^{5 t} + c_{2} - 2 e^{4 t} - \frac{2}{3} e^{- t}

Solution steps

y^{''} (t) - 5 y^{'} (t) = 8 e^{4 t} - 4 e^{- t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} - 2 e^{4 t} - \frac{2}{3} e^{- t}

y = c_{1} e^{5 t} + c_{2} - 2 e^{4 t} - \frac{2}{3} e^{- t}

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

y^{^{''}} + 3 y^{^{'}} = 18

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d x ^{2}} + y = 4 x

y^{^{''}} + 5 y^{^{'}} + 6 y = 1 + e^{- t}

What is the solution for y^{''}-5y^'=8e^{4t}-4e^{-t} ?
The solution for y^{''}-5y^'=8e^{4t}-4e^{-t} is y=c_{1}e^{5t}+c_{2}-2e^{4t}-2/3 e^{-t}