If $\phi(x) = \frac{1}{\sqrt{x}} \int \limits_0^x (4 \sqrt{2} \sin t - 3 \phi^{\prime}(t)) dt, \quad x > 0$,then $\phi^{\prime}\left(\frac{\pi}{4}\right)$ is equal to:

The curve that satisfies the differential equation $x y \, dy - (1 + y^2) \, dx = 0$ passes through $(1, 0)$ and intersects the curve $x^2 + 3y^2 = 3$ at an angle $\theta$. Then $\frac{2\theta}{\pi} =$

The solution of the differential equation $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$ satisfying $y = 1$ when $x = 1$ 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 $f:[0, \infty) \rightarrow R$ 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 $\frac{\pi-1}{4}$

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$