The solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ is

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ at $x=1, y=0$,represents

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x)f'(y)+f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\sum_{n=1}^{100} \log_{e} f(n)$ is equal to:

If the solution $y(x)$ of the given differential equation $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ passes through the point $(\frac{\pi}{2}, 0)$,then the value of $e^{y(\frac{\pi}{6})}$ is equal to ...........

If $f(x)$ is a function such that $f^{\prime}(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$,then $f(1)=$

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation $y(1 + xy)dx = xdy$,then $f(-\frac{1}{2}) = $ . . . . .