What is the x'(t)= 1/(3x^2+1),x(0)=1 ?

The x'(t)= 1/(3x^2+1),x(0)=1 is 1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x'(t)= 1/(3x^2+1),x(0)=1

x^{'} (t) = \frac{1}{3 x ^{2} + 1}, x (0) = 1

\frac{1}{4}

Solution steps

\frac{1}{3 x ^{2} + 1} = \frac{1}{3 x ^{2} + 1}

Constant function results in a constant

x (0) = 1

= \frac{1}{3 \cdot 1 ^{2} + 1}

\frac{1}{3 \cdot 1 ^{2} + 1} = \frac{1}{4}

= \frac{1}{4}

\frac{\partial}{\partial x} (e^{- 6 t} cos (π x))

\frac{\partial}{\partial y} (9 z e^{x yz})

\int_{0}^{1} (e^{- 0.1 t}) d t

\int_{0}^{π} 2 sin (n x) d x

f (x) = 14 x