What is the solution for x^{''}+10x^'+25=14e^{-5t} ?

The solution for x^{''}+10x^'+25=14e^{-5t} is x=c_{1}+c_{2}e^{-10t}-14/25 e^{-5t}-(5t)/2

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x^{''}+10x^'+25=14e^{-5t}

x^{^{''}} + 10 x^{^{'}} + 25 = 14 e^{- 5 t}

x = c_{1} + c_{2} e^{- 10 t} - \frac{14}{25} e^{- 5 t} - \frac{5 t}{2}

Solution steps

x^{''} (t) + 10 x^{'} (t) + 25 = 14 e^{- 5 t}

Solve linear ODE:

x = c_{1} + c_{2} e^{- 10 t} - \frac{14}{25} e^{- 5 t} - \frac{5 t}{2}

x = c_{1} + c_{2} e^{- 10 t} - \frac{14}{25} e^{- 5 t} - \frac{5 t}{2}

\frac{d y}{d t} = - 4 (y - 3)

x^{4} - 50 x^{2}

\frac{x ^{2}}{x + 1}

x \to 2 lim (9 - x)

\int \frac{2 u}{u + 2} d u

What is the solution for x^{''}+10x^'+25=14e^{-5t} ?
The solution for x^{''}+10x^'+25=14e^{-5t} is x=c_{1}+c_{2}e^{-10t}-14/25 e^{-5t}-(5t)/2