Let $y$ be the solution of the differential equation $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$ satisfying $y(1) = 1$. Then,$y$ satisfies:

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

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,$4\theta - \frac{\pi}{2} =$

Let $b$ be a nonzero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=1$. If the derivative $f^{\prime}$ of $f$ satisfies the equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$ for all $x \in R$,then which of the following statements is/are $TRUE$?
$(A)$ If $b>0$,then $f$ is an increasing function
$(B)$ If $b < 0$,then $f$ is a decreasing function
$(C)$ $f(x)f(-x)=1$ for all $x \in R$
$(D)$ $f(x)-f(-x)=0$ for all $x \in R$

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 $x$):

Find the family of curves such that the angle between the tangent at any point $(x, y)$ and the tangent to the curve $xy = c^2$ at the point of intersection is $\frac{\pi}{4}$.