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

The solution for y^{''}+(8)y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+(8)y=0

y^{^{''}} + (8) y = 0

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

Solution steps

y^{''} (t) + (8) y = 0

Solve linear ODE:

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

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

h \to + 0 lim (\frac{( 2 ^{h} - 1 )}{h})

n = 0 \sum \infty 100

y^{^{''}} - 2 y^{^{'}} + y = x^{2}

\frac{\partial}{\partial x} (x^{3} - 5 x y)

\frac{x d y}{d x} + (3 x + 1) y = e^{- 3 x}

What is the solution for y^{''}+(8)y=0 ?
The solution for y^{''}+(8)y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)