The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be:

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 $\frac{d^2y}{dx^2} = \cos x - \sin x$ is

Verify that the given function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$.

$A$ function $y = f(x)$ satisfies the differential equation $\frac{dy}{dx} - y = \cos x - \sin x$ with the initial condition that $y$ is bounded when $x \rightarrow \infty$. The area enclosed by $y = f(x)$,$y = \cos x$,and the $y$-axis is

If $y = \frac{x}{\ln |c x|}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,then the function $\phi \left( \frac{x}{y} \right)$ is: