The solution of $\frac{dy}{dx} = e^{-2x}$ with the condition $y(\log 2) = \frac{1}{16}$ is $y =$

$A$ 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:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then the value of $(2y(1) - 1)$ is equal to:

The general solution of the differential equation $\frac{x dy - y dx}{y} = 0$ is . . . . . . .

If a curve passes through $(1, 2)$ and has the slope of its tangent $1 - \frac{1}{x^2}$ at any point $(x, y)$,then the equation of that curve is:

The particular solution of $\frac{dy}{dx} = 1 + x + y^2 + xy^2$,when $y(0) = 0$,is