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

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

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

y = \frac{e ^{2 t} ( 2 - c _{2} )}{2} + c_{2} e^{t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{2 t} ( 2 - c _{2} )}{2} + c_{2} e^{t}

y = \frac{e ^{2 t} ( 2 - c _{2} )}{2} + c_{2} e^{t}

y^{^{''''}} + 6 y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{'''}} + 2 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

y^{^{''}} - 8 y^{^{'}} - 12 y = 0

y^{^{''''}} + 2 y^{^{'''}} + 10 y^{^{''}} + 18 y^{^{'}} + 9 y = 0

3 y^{^{''}} - 5 y^{^{'}} = 0

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