What is the (dy)/(dx)=((x^2+1))/(y^2+1),y(0)=1 ?

The (dy)/(dx)=((x^2+1))/(y^2+1),y(0)=1 is (x^2+1)/2

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(dy)/(dx)=((x^2+1))/(y^2+1),y(0)=1

\frac{d y}{d x} = \frac{( x ^{2} + 1 )}{y ^{2} + 1}, y (0) = 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 (0) = 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}

\frac{d}{d x} (arccos (1 - 2 x^{2}))

x \to 3 lim (\frac{3 x ^{2} - 1 - 2}{x - 3})

y = 6 - 6 x, y = 0, x = 0

\frac{\partial}{\partial x} (- x^{2} + 2 x y - y^{2})

x \to - 3 lim (2 x + ∣ x - 3 ∣)

What is the (dy)/(dx)=((x^2+1))/(y^2+1),y(0)=1 ?
The (dy)/(dx)=((x^2+1))/(y^2+1),y(0)=1 is (x^2+1)/2