What is the solution for y^{''}+4y^'+34y=0 ?

The solution for y^{''}+4y^'+34y=0 is y=e^{-2t}(c_{1}cos(sqrt(30)t)+c_{2}sin(sqrt(30)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+34y=0

y^{^{''}} + 4 y^{^{'}} + 34 y = 0

y = e^{- 2 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

y = e^{- 2 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

f (x) = \frac{2}{4 x + 3}

\int_{1}^{e^{2}} \frac{ln ^{6} ( x ^{2} )}{x} d x

f (x) = \frac{x - 4}{x + 2}

\frac{x}{36 + x ^{2}}

x^{2} y d y - x y^{2} d x - x^{3} y^{2} d x = 0

What is the solution for y^{''}+4y^'+34y=0 ?
The solution for y^{''}+4y^'+34y=0 is y=e^{-2t}(c_{1}cos(sqrt(30)t)+c_{2}sin(sqrt(30)t))