$A$ solution of the differential equation ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$ is

The solution of the differential equation $x dy = (y + xy^3 (1 + \log_e x)) dx$ is (where $C$ is an arbitrary constant):

Solve the differential equation $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)\,dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y\,dy = 0$.

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:

$A$ continuously differentiable function $\phi (x)$ in $(0, \pi)$ satisfying $y' = 1 + y^2$ and $y(0) = 0 = y(\pi)$ is

The general solution of $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$ is