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

The solution for x^{''}+x^'=e^{-t} is x=c_{1}+c_{2}e^{-t}-e^{-t}t

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x^{''}+x^'=e^{-t}

x^{^{''}} + x^{^{'}} = e^{- t}

x = c_{1} + c_{2} e^{- t} - e^{- t} t

Solution steps

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

Solve linear ODE:

x = c_{1} + c_{2} e^{- t} - e^{- t} t

x = c_{1} + c_{2} e^{- t} - e^{- t} t

y^{^{''}} + 49 y = 8 cos (7 x) (3 x + 2)

y^{^{''}} + y^{^{'}} - 6 y = t - 2 π, y (0) = 1, y^{^{'}} (0) = 2

- v y^{^{''}} - a y = c

y^{^{''}} - 7 y^{^{'}} + 10 y = e^{3 x} (- 1 + 10 x)

(d^{3} - 3 d - 2) y = 540 x^{3} e - x

What is the solution for x^{''}+x^'=e^{-t} ?
The solution for x^{''}+x^'=e^{-t} is x=c_{1}+c_{2}e^{-t}-e^{-t}t