What is the solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=2 is y=e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'-6y=0,y(0)=1,y^'(0)=2

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{2 t}

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = 0

Solve linear ODE:

y = e^{2 t}

y = e^{2 t}

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 3

x^{^{''}} + 5 x^{^{'}} + 6 x = 0

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 1

y^{^{''''}} + 14 y^{^{''}} + 49 y = 0

What is the solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=2 is y=e^{2t}