The solution of $e^{dy/dx} = x$ with the initial conditions $x = 1$ and $y = 0$ is:

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$,then a value of $\alpha$ is

Let $x = x(y)$ be the solution of the differential equation $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$,$y > 0$ and $x(1) = \frac{\pi}{2}$. Then $\cos(x(2))$ is equal to:

The general solution of the differential equation $y(1+\log x)\left(\frac{dx}{dy}\right) - x \log x = 0$ is

If the slope of the tangent at a point $(x, y)$ on a curve is $\frac{y-4}{x-3}$ and the curve passes through $(4, 3)$,then the point where it cuts the line $y=x$ is

The solution of $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ is