The general solution of $\frac{dy}{dx} = 2xye^{x^2}$ 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:

The solution of $\log \left(\frac{dy}{dx}\right) = 2x - 5y$ with the initial condition $y(0) = 0$ is:

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx}+e^x=0$ satisfying $y(0)=1$,then a value of $y(\log 13)$ is

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

Let $y=y(x)$ be a solution of the differential equation,$\sqrt{1-x^{2}} \frac{dy}{dx}+\sqrt{1-y^{2}}=0, |x| < 1$. If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$,then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to