What is the solution for y^{''}+y^'+1/3 y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'+1/3 y=0

y^{^{''}} + y^{^{'}} + \frac{1}{3} y = 0

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

Solution steps

y^{''} (t) + y^{'} (t) + \frac{1}{3} y = 0

Solve linear ODE:

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

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

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

x^{^{''}} + 3 x^{^{'}} - 3 x = 0

y^{^{''}} + 8 y^{^{'}} + 160 y = 0

0.1 y^{^{''}} + y^{^{'}} + y = 0, y (0) = 1, y (1) = 0

y^{^{''}} + 0.2 y^{^{'}} + 0.01 y = 0

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