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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

y = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

y^{^{''}} + y^{^{'}} + 3 y = 0

y^{^{''}} + 225 y = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

2 y^{^{''}} + 2 y^{^{'}} + 3 y = 0

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

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