The curve which passes through the point $(\sqrt{2}, 1)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{2x}{3y}$ represents:

The solution of $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$ is

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

Find the particular solution of the differential equation $(1+e^{x}) dy+(1+y^{2}) e^{x} dx=0$,given that $y=1$ when $x=0$.

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}) = $ . . . . .

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: