What is the solution for y^{''}-6y^'+13y=0,y(0)=-3 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-6y^'+13y=0,y(0)=-3

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = - 3

y = e^{3 t} (- 3 cos (2 t) + c_{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{3 t} (- 3 cos (2 t) + c_{2} sin (2 t))

y = e^{3 t} (- 3 cos (2 t) + c_{2} sin (2 t))

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