What is the solution for x^{''}-kx=0 ?

The solution for x^{''}-kx=0 is x=c_{1}e^{sqrt(k)t}+c_{2}e^{-sqrt(k)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-kx=0

x^{^{''}} - k x = 0

x = c_{1} e^{k t} + c_{2} e^{- k t}

Solution steps

x^{''} (t) - k x (t) = 0

Solve linear ODE:

x = c_{1} e^{k t} + c_{2} e^{- k t}

x = c_{1} e^{k t} + c_{2} e^{- k t}

\frac{\partial}{\partial t} (e^{- 2 t} cos (π x))

\int (1 3^{x}) d x

x \to - 2 lim (3 x^{3} + 6 x^{2} + 4)

5 x \cdot \frac{d y}{d x} = 2 x e^{x} - 5 y + 6 x^{2}

\int 3 x - x d x

What is the solution for x^{''}-kx=0 ?
The solution for x^{''}-kx=0 is x=c_{1}e^{sqrt(k)t}+c_{2}e^{-sqrt(k)t}