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 $f:[0, \infty) \rightarrow R$ be a continuous function such that $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ for all $x \in[0, \infty)$. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$
$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$
$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$
$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$

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

The solution of $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$ is

The differential equation $\frac{dx}{dy} = \frac{3y}{2x}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity:

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is