What is the solution for (d^2x)/(dt^2)+10(dx)/(dt)+25x=0 ?

The solution for (d^2x)/(dt^2)+10(dx)/(dt)+25x=0 is x=c_{1}e^{-5t}+c_{2}te^{-5t}

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(d^2x)/(dt^2)+10(dx)/(dt)+25x=0

\frac{d ^{2} x}{d t ^{2}} + 10 \frac{d x}{d t} + 25 x = 0

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 10 \frac{d x}{d t} + 25 x = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 9, y^{^{'}} (0) = - 60

y^{^{'''}} - 7 y^{^{''}} + 12 y^{^{'}} = 0

x^{^{''}} + 2 x^{^{'}} - 8 x = 0, x^{^{'}} (0) = 22, x (0) = - 1

\frac{d ^{2} x}{d t ^{2}} - 4 \frac{d x}{d t} + 5 x = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y^{1} (0) = e^{x} cos (2 x)

What is the solution for (d^2x)/(dt^2)+10(dx)/(dt)+25x=0 ?
The solution for (d^2x)/(dt^2)+10(dx)/(dt)+25x=0 is x=c_{1}e^{-5t}+c_{2}te^{-5t}