What is the solution for y^{''}-6y^'+18y=0,y(0)=0,y^'(0)=6 ?

The solution for y^{''}-6y^'+18y=0,y(0)=0,y^'(0)=6 is y=2e^{3t}sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+18y=0,y(0)=0,y^'(0)=6

y^{^{''}} - 6 y^{^{'}} + 18 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y = 2 e^{3 t} sin (3 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = 2 e^{3 t} sin (3 t)

y = 2 e^{3 t} sin (3 t)

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

y^{^{''}} + \frac{64}{5} y^{^{'}} + 64 y = 0

25 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 8

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0

What is the solution for y^{''}-6y^'+18y=0,y(0)=0,y^'(0)=6 ?
The solution for y^{''}-6y^'+18y=0,y(0)=0,y^'(0)=6 is y=2e^{3t}sin(3t)