$A$ curve $y = f(x)$ passing through the point $\left(1, \frac{1}{\sqrt{e}}\right)$ satisfies the differential equation $\frac{dy}{dx} + x e^{-\frac{x^2}{2}} = 0.$ Then which of the following does not hold good?

Which one of the following curves represents the solution of the initial value problem $Dy = 100 - y$,where $y(0) = 50$?

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

The solution of $(1+xy)y \, dx + (1-xy)x \, dy = 0$ is:

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 $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ is