Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by $5$ is

Let $y=y(x)$ be a solution curve of the differential equation $(y+1) \tan ^{2} x \,dx+\tan x \,dy+y \,dx=0$ for $x \in \left(0, \frac{\pi}{2}\right)$. If $\lim _{x \rightarrow 0+} x y(x)=1$,then the value of $y\left(\frac{\pi}{4}\right)$ is:

If $\frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x}$,$x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ and $y\left( \frac{\pi}{4} \right) = \frac{4}{3}$,then $y\left( -\frac{\pi}{4} \right)$ equals

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to: