What is the solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 is y=(1-8/13)e^{7t}+8/13 e^{-6t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1

y^{^{''}} - y^{^{'}} - 42 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

Solution steps

y^{''} (t) - y^{'} (t) - 42 y = 0

Solve linear ODE:

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

What is the solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 is y=(1-8/13)e^{7t}+8/13 e^{-6t}