If the general solution of the differential equation $(y-x+1) dy - (y+x+2) dx = 0$ is $f(x, y, c) = 0$, then the value of $c$ such that $f(1, 1, c) = 0$ is

The slope of the normal at any point $(x, y), x > 0, y > 0$ on the curve $y=y(x)$ is given by $\frac{x^{2}}{x y-x^{2} y^{2}-1}$. If the curve passes through the point $(1, 1)$,then $e \cdot y(e)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 dx=dy$,$y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ be $\alpha$ and $\beta$,respectively. If $(3\alpha+\pi)^2+\beta^2=\gamma+\delta\sqrt{3}$,where $\gamma, \delta \in Z$,then $\gamma+\delta$ equals:

If $2xy^3dx + x^2y^2dy = ydx - xdy$ and $y(2) = 1$,then the value of $y(-1)$ will be (where $y(x)$ denotes the value of $y$ for a given $x$):

The solution of the differential equation $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$ is:

The solution of $\frac{d^2y}{dx^2} = \sec^2 x + x e^x$ is