The solution of the differential equation $(1+x) y \,dx + (1-y) x \,dy = 0$ is

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{dx}{dt}+ax=0$ and $\frac{dy}{dt}+by=0$ respectively,where $a, b \in R$. Given that $x(0)=2$,$y(0)=1$,and $3y(1)=2x(1)$,the value of $t$ for which $x(t)=y(t)$ is:

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ is

If $\frac{dy}{dx} = \frac{xy + y}{xy + x}$,then the solution of the differential equation is:

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$,then $f(xy)$ is equal to