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

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

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

y = e^{- t + 1} t

Solution steps

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

Solve linear ODE:

y = e^{- t + 1} t

y = e^{- t + 1} t

\frac{d}{d x} (x^{4} - 4 x^{3} + 3 x^{2} - 3)

\int (x^{2} - 2 x + 5) e^{- x} d x

x \to - 2 lim (7 x^{2} + 8)

cos (- 3 x)

sin (t) - sin (t) cos (t) - 5 t

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