What is the solution for y^{''}-4y^'-5y=2+e^{-t} ?

The solution for y^{''}-4y^'-5y=2+e^{-t} is y=c_{1}e^{5t}+c_{2}e^{-t}-2/5-(e^{-t}t)/6

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y^{''}-4y^'-5y=2+e^{-t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 + e^{- t}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{5} - \frac{e ^{- t} t}{6}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 2 + e^{- t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{5} - \frac{e ^{- t} t}{6}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{5} - \frac{e ^{- t} t}{6}

y^{^{''}} + 16 y = e^{- 4 t} + 3 sin (4 t)

\frac{d ^{2} y}{d x ^{2}} + y = 5 sin (4 x)

y^{^{''}} - 4 y = - 6 cos (4 x) + 2 sin (4 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{( 5 + e ^{x} )}

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-4y^'-5y=2+e^{-t} ?
The solution for y^{''}-4y^'-5y=2+e^{-t} is y=c_{1}e^{5t}+c_{2}e^{-t}-2/5-(e^{-t}t)/6