What is the solution for y^{''}-3y^'+2y=0,y(0)=4,y^'(0)=2 ?

The solution for y^{''}-3y^'+2y=0,y(0)=4,y^'(0)=2 is y=-2e^{2t}+6e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = - 2 e^{2 t} + 6 e^{t}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = - 2 e^{2 t} + 6 e^{t}

y = - 2 e^{2 t} + 6 e^{t}

y^{^{''}} + \frac{a}{4} y = 0

y^{^{''}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y^{^{''}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{'''}} + 8 y^{^{''}} + 21 y^{^{'}} + 18 y = 0

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

What is the solution for y^{''}-3y^'+2y=0,y(0)=4,y^'(0)=2 ?
The solution for y^{''}-3y^'+2y=0,y(0)=4,y^'(0)=2 is y=-2e^{2t}+6e^t