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

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

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(d^2y)/(dt^2)+10(dy)/(dt)+25y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{x ^{3} y + x y ^{3}}{x ^{2} + y ^{2}})

f (x) = x - 4, (13, 3)

f (x) = \frac{5}{x + 1}, at x = \frac{- 1}{3}

x \to 0 lim (\frac{0 + 5}{3 ( 0 )})

f (x) = \frac{x ^{2} + 8 x + 3}{x}

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