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

The solution of the differential equation $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ at $x=0, y=1$ is

$A$ function $y = f(x)$ satisfies the condition $f'(x) \sin x + f(x) \cos x = 1$,where $f(x)$ is bounded as $x \rightarrow 0$. If $I = \int_{0}^{\frac{\pi}{2}} f(x) \, dx$,then:

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

If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y)dx + xdy = 0$ passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$,then $|y(1)|$ 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 $\frac{2\theta}{\pi} =$