What is the solution for x^{''}+3x^'+2x=e^{-t} ?

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

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x^{''}+3x^'+2x=e^{-t}

x^{^{''}} + 3 x^{^{'}} + 2 x = e^{- t}

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

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = e^{- t}

Solve linear ODE:

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

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

(3 x + 25)

\frac{\partial}{\partial x} (e^{2 x} cos (y))

x \to - 2 lim (3 x + 4)

\int x e^{5 x} d x

h \to 0 lim (\frac{1}{h ^{2}})

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