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

The solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+1/3-1/3 e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'=0,y(0)=1,y^'(0)=1

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

3 x^{\frac{3}{2}}

(2, 3), (7, - 6)

\frac{d}{d x} (arctan (\frac{x}{4}))

f (x) = \frac{1}{x + 1}, (0, 1)

What is the solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+1/3-1/3 e^{-3t}