What is the solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=2 is y=e^{-2t}(cos(t)+4sin(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- 2 t} (cos (t) + 4 sin (t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- 2 t} (cos (t) + 4 sin (t))

y = e^{- 2 t} (cos (t) + 4 sin (t))

\int \frac{1}{1 - 3 v} d v

x \to 9 lim (\frac{x ^{2}}{3} - \frac{3}{x})

\int 4 sin^{2} (x) d x

\int_{2}^{5} 1 d x

\frac{d}{d x} (\frac{x}{x + 13})

What is the solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=2 is y=e^{-2t}(cos(t)+4sin(t))