What is the solution for y^{''}+3y=0,y(0)= pi/8 ,y^'(0)=0 ?

The solution for y^{''}+3y=0,y(0)= pi/8 ,y^'(0)=0 is y= pi/8 cos(sqrt(3)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y=0,y(0)= pi/8 ,y^'(0)=0

y^{^{''}} + 3 y = 0, y (0) = \frac{π}{8}, y^{^{'}} (0) = 0

y = \frac{π}{8} cos (3 t)

Solution steps

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

Solve linear ODE:

y = \frac{π}{8} cos (3 t)

y = \frac{π}{8} cos (3 t)

4 y^{^{''}} + 25 y = 0

3 y^{^{''}} - 12 y^{^{'}} + 7 y = 0, y (1) = e^{2 x} sin (2 x)

y^{^{''}} - 4 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = - 1

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

x^{^{''}} + \frac{15}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

What is the solution for y^{''}+3y=0,y(0)= pi/8 ,y^'(0)=0 ?
The solution for y^{''}+3y=0,y(0)= pi/8 ,y^'(0)=0 is y= pi/8 cos(sqrt(3)t)