What is the solution for y^{''}-8y=0 ?

The solution for y^{''}-8y=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-8y=0

y^{^{''}} - 8 y = 0

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t}

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t}

a x^{^{''}} + b x = 0

\frac{d}{d x} (e^{t a n h (2 x)})

\int 1 - cos (6 x) d x

(0.5) (1.6)

\frac{d y}{d x} x^{3} + y^{3} = 6 x y

What is the solution for y^{''}-8y=0 ?
The solution for y^{''}-8y=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}