What is the solution for y^{''}+y^'+6y=0 ?

The solution for y^{''}+y^'+6y=0 is y=e^{-t/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'+6y=0

y^{^{''}} + y^{^{'}} + 6 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

Solution steps

y^{''} (t) + y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{8 + e ^{x}}

(4^{x})^{9}

\frac{d y}{d x}, 4^{x} = 2 x y

\int \frac{x}{x + 4} d x

(x + y) d x + (x - y) d y = 0

What is the solution for y^{''}+y^'+6y=0 ?
The solution for y^{''}+y^'+6y=0 is y=e^{-t/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))