What is the solution for 3/16 y^{''}+12y=0,y(0)= 2/3 ,y^'(0)=-3 ?

The solution for 3/16 y^{''}+12y=0,y(0)= 2/3 ,y^'(0)=-3 is y= 2/3 cos(8t)-3/8 sin(8t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3/16 y^{''}+12y=0,y(0)= 2/3 ,y^'(0)=-3

\frac{3}{16} y^{^{''}} + 12 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - 3

y = \frac{2}{3} cos (8 t) - \frac{3}{8} sin (8 t)

Solution steps

\frac{3}{16} y^{''} (t) + 12 y = 0

Solve linear ODE:

y = \frac{2}{3} cos (8 t) - \frac{3}{8} sin (8 t)

y = \frac{2}{3} cos (8 t) - \frac{3}{8} sin (8 t)

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + (a^{2} - 2) y = 0

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

25 x^{^{''}} + 5 x^{^{'}} + 204.167 x = 0, x (0) = 28.577

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

What is the solution for 3/16 y^{''}+12y=0,y(0)= 2/3 ,y^'(0)=-3 ?
The solution for 3/16 y^{''}+12y=0,y(0)= 2/3 ,y^'(0)=-3 is y= 2/3 cos(8t)-3/8 sin(8t)