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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 9 x

y^{^{''}} + 9 y^{^{'}} + 22 y = e^{- 3 x}

(D^{2} - 4 D + 5) y = (x + 1)^{3}

d, d (t) = \frac{1}{2} t^{2} - 10 t

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 7 + x^{- 1} e^{x}

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