What is the solution for 9y^{''}+6y=0 ?

The solution for 9y^{''}+6y=0 is y=c_{1}cos(sqrt(2/3)t)+c_{2}sin(sqrt(2/3)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9y^{''}+6y=0

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

y = c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t)

y = c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t)

\int \frac{- cos ( 2 x ) - 1}{4} d x

\int \frac{3}{98} x^{2} d x

f (x) = cos^{2} (3 x)

y^{^{'}} = 8 x^{3} \cdot y^{2}

x \to 0 lim (\frac{sin ( 7 x )}{tan ( 5 x )})

What is the solution for 9y^{''}+6y=0 ?
The solution for 9y^{''}+6y=0 is y=c_{1}cos(sqrt(2/3)t)+c_{2}sin(sqrt(2/3)t)