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

The solution for x^{''}+x^'-2x=e^t is x=c_{1}e^t+c_{2}e^{-2t}+(e^tt)/3

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

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

x = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{t} t}{3}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{t} t}{3}

x = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{t} t}{3}

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\frac{\partial}{\partial x} (e^{4 x y} \cdot 4 y)

What is the solution for x^{''}+x^'-2x=e^t ?
The solution for x^{''}+x^'-2x=e^t is x=c_{1}e^t+c_{2}e^{-2t}+(e^tt)/3