Let $y$ be the solution of the differential equation $(1-x^{2}) dy = (xy + (x^{3}+2) \sqrt{1-x^{2}}) dx$ for $-1 < x < 1$ with $y(0)=0$. If $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) dx = k$,then $k^{-1}$ is equal to:

If for the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}$,$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$,$f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$,then $f\left(\frac{\pi}{4}\right)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$,where $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$,then $y(\frac{\pi}{6})$ is:

Let $\alpha$ be a non-zero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=2$ and $\lim _{x \rightarrow-\infty} f(x)=1$. If $f^{\prime}(x)=\alpha f(x)+3$ for all $x \in R$,then $f(-\log _e 2)$ is equal to . . . . . . . . .

Find the general solution of the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$ where $y > 0$.

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........