If $\frac{dy}{dx} = \frac{2^x y + 2^y \cdot 2^x}{2^x + 2^{x+y} \log_e 2}$ and $y(0) = 0$,then for $y = 1$,the value of $x$ lies in the interval:

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 $\cos y \frac{dy}{dx} = e^{x+\sin y} + x^2 e^{\sin y}$ is $f(x) + e^{-\sin y} = C$ ($C$ is an arbitrary real constant), where $f(x)$ is equal to:

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

The solution of the differential equation $(1 - x^2)dy + xydx = xy^2dx$ is

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