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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = (1 - c_{2}) e^{t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{t} + c_{2} e^{- t}

y = (1 - c_{2}) e^{t} + c_{2} e^{- t}

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = - 4, y^{^{'}} (0) = 3

x^{^{''}} + 4 x^{^{'}} - 5 x = 0

9 y^{^{''}} - 42 y^{^{'}} + 49 y = 0

x^{^{''}} + 2 a x^{^{'}} + b x = 0

y^{^{''''}} + 4 y^{^{'''}} + 4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

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