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 $\frac{\pi}{4}$.

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

Let the tangent at any point $P(x, y)$ on a curve passing through the points $(1, 1)$ and $(\frac{1}{10}, 100)$ intersect the positive $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If $PA: PB = 1: k$ and $y = y(x)$ is the solution of the differential equation $e^{\frac{dy}{dx}} = 2x + 1$ with $y(0) = 2$,then $4y(1) - 5 \log_e 3$ is equal to:

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

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 $\frac{2\theta}{\pi} =$

If $f(x), f^{\prime}(x), f^{\prime \prime}(x)$ are positive functions and $f(0)=1, f^{\prime}(0)=2$,then the solution of the differential equation $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ is