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

The solution for y^{''}+2y^'+y=0,y(0)=1,y^'(0)=2 is y=e^{-t}+3e^{-t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- t} + 3 e^{- t} t

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- t} + 3 e^{- t} t

y = e^{- t} + 3 e^{- t} t

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 14

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0

x^{^{''}} - 6 x^{^{'}} + 9 x = 0

y^{^{''}} + 144 y = 0

What is the solution for y^{''}+2y^'+y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+2y^'+y=0,y(0)=1,y^'(0)=2 is y=e^{-t}+3e^{-t}t