What is the (dy)/(dx)=(x^2-1)/(y^2+1),y(-1)=1 ?

The (dy)/(dx)=(x^2-1)/(y^2+1),y(-1)=1 is (x^2-1)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(x^2-1)/(y^2+1),y(-1)=1

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2} + 1}, y (- 1) = 1

\frac{x ^{2} - 1}{2}

Solution steps

\frac{x ^{2} - 1}{y ^{2} + 1} = \frac{x ^{2} - 1}{y ^{2} + 1}

Constant function results in a constant

y (- 1) = 1

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

Simplify

\frac{x ^{2} - 1}{1 ^{2} + 1} : \frac{x ^{2} - 1}{2}

= \frac{x ^{2} - 1}{2}

n \to \infty lim (\frac{n}{- n})

x \to 3 lim (\frac{3 x - 5}{3 - x})

\frac{d ^{2} x}{d t ^{2}} = - k \cdot x

\int 9 sec^{2} (x) d x

x + ln (x)

What is the (dy)/(dx)=(x^2-1)/(y^2+1),y(-1)=1 ?
The (dy)/(dx)=(x^2-1)/(y^2+1),y(-1)=1 is (x^2-1)/2