What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-7 ?

The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{3t}sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-7

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

y = - \frac{7}{2} e^{3 t} sin (2 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = - \frac{7}{2} e^{3 t} sin (2 t)

y = - \frac{7}{2} e^{3 t} sin (2 t)

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

y^{^{'''}} + 4 y^{^{''}} - 4 y^{^{'}} - 16 y = 0

y^{^{''}} = 10 y + 19 y^{^{'}}

y^{^{''''}} - y^{^{'''}} - 3 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

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

What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-7 ?
The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{3t}sin(2t)