What is the solution for y^{''}+4y^'+4=0,y^'(0)=1 ?

The solution for y^{''}+4y^'+4=0,y^'(0)=1 is y=c_{1}-1/2 e^{-4t}-t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4=0,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 4 = 0, y^{^{'}} (0) = 1

y = c_{1} - \frac{1}{2} e^{- 4 t} - t

Solution steps

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

Solve linear ODE:

y = c_{1} - \frac{1}{2} e^{- 4 t} - t

y = c_{1} - \frac{1}{2} e^{- 4 t} - t

x \to \infty lim (- \frac{1}{x})

f (x) = cos (x), - \frac{π}{2}, \frac{π}{2}

x y^{^{'}} = y + y^{3}

\frac{d y}{d x} + \frac{y}{x} = x y^{2}

x \to 7 lim (\frac{1}{∣ 7 - x ∣})

What is the solution for y^{''}+4y^'+4=0,y^'(0)=1 ?
The solution for y^{''}+4y^'+4=0,y^'(0)=1 is y=c_{1}-1/2 e^{-4t}-t