What is the solution for 16y^{''}+3y=0 ?

The solution for 16y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/4)+c_{2}sin((sqrt(3)t)/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16y^{''}+3y=0

16 y^{^{''}} + 3 y = 0

y = c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})

y = c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})

y^{^{''}} + 20 y^{^{'}} + 200 = 300

\int_{0}^{N} e^{- s t} sin (2 t) d t

\frac{1}{1 + 4 t ^{2}}

\frac{4 x}{x + 1} (3.3)

y = t (ln (3 t))^{2}

What is the solution for 16y^{''}+3y=0 ?
The solution for 16y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/4)+c_{2}sin((sqrt(3)t)/4)