What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.45 ?

The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.45 is y=4e^{6t}-8.55e^{6t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.45

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.45

y = 4 e^{6 t} - 8.55 e^{6 t} t

Solution steps

y^{''} (t) - 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 4 e^{6 t} - 8.55 e^{6 t} t

y = 4 e^{6 t} - 8.55 e^{6 t} t

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.46

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

y^{^{''''}} + 5 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0

y + 2 y^{^{'}} + y^{^{''}} = 0

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

What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.45 ?
The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.45 is y=4e^{6t}-8.55e^{6t}t