What is the solution for y^{''}+2y^'+10y=0 ?

The solution for y^{''}+2y^'+10y=0 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+10y=0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

f (x) = arcsin (5 x)

x \to 5 lim (\frac{x - 5}{2 x - 10})

x \to - 1 lim (\frac{x ^{2} - 9}{x + 1})

n = 2 \sum \infty \frac{2}{n ^{2} + 1}

\int (7 - 3 w - 9 w^{2}) d w

What is the solution for y^{''}+2y^'+10y=0 ?
The solution for y^{''}+2y^'+10y=0 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))