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:

The solution of the differential equation $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$ is (where $c$ is the constant of integration):

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:

If a curve passes through the point $\left( 2, \frac{7}{2} \right)$ and has slope $\left( 1 - \frac{1}{x^2} \right)$ at any point $(x, y)$ on it,then the ordinate of the point on the curve whose abscissa is $-2$ is

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

The solution of the differential equation $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$ is