What is the solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=1 is y=e^{3t}-e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{3 t} - e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{3 t} - e^{2 t}

y = e^{3 t} - e^{2 t}

2 y^{^{''}} - 12 y^{^{'}} + 18 y = 0

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

y^{^{''}} + 40 y^{^{'}} + 6250 y = 0

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

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

What is the solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=1 is y=e^{3t}-e^{2t}