What is the solution for 3/4 x^{''}+9x=0 ?

The solution for 3/4 x^{''}+9x=0 is x=c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3/4 x^{''}+9x=0

\frac{3}{4} x^{^{''}} + 9 x = 0

x = c_{1} cos (23 t) + c_{2} sin (23 t)

Solution steps

\frac{3}{4} x^{''} (t) + 9 x = 0

Solve linear ODE:

x = c_{1} cos (23 t) + c_{2} sin (23 t)

x = c_{1} cos (23 t) + c_{2} sin (23 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 6, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 35 y = 0

(4 D^{3} - 15 D^{2} + 5 D + 6) y = 0

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

y^{^{''}} + 5 y^{^{'}} - 24 y = 0, y (0) = a, y^{^{'}} (0) = 80

What is the solution for 3/4 x^{''}+9x=0 ?
The solution for 3/4 x^{''}+9x=0 is x=c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t)