What is the solution for y^{''}+7y^'+23y=0 ?

The solution for y^{''}+7y^'+23y=0 is y=e^{-(7t)/2}(c_{1}cos((sqrt(43)t)/2)+c_{2}sin((sqrt(43)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+7y^'+23y=0

y^{^{''}} + 7 y^{^{'}} + 23 y = 0

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{43 t}{2}) + c_{2} sin (\frac{43 t}{2}))

Solution steps

y^{''} (t) + 7 y^{'} (t) + 23 y = 0

Solve linear ODE:

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{43 t}{2}) + c_{2} sin (\frac{43 t}{2}))

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{43 t}{2}) + c_{2} sin (\frac{43 t}{2}))

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

\int \frac{2 z - 3}{7 z - 10} d z

\frac{\partial}{\partial x} (2 sin (x) cos (y))

\frac{d}{d x} (\frac{x}{x ^{- 1} + 1})

\frac{\partial}{\partial x} (x^{3} y + 12 x^{2} - 64 y)

What is the solution for y^{''}+7y^'+23y=0 ?
The solution for y^{''}+7y^'+23y=0 is y=e^{-(7t)/2}(c_{1}cos((sqrt(43)t)/2)+c_{2}sin((sqrt(43)t)/2))