The solution of the differential equation $(x + \log y)dy + y\,dx = 0$ is:

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

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{(x^2-1)^{5/2}}$ for $x > 1$,such that $y(2)=\frac{2}{9}\log_e(2+\sqrt{3})$ and $y(\sqrt{2})=\alpha\log_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}$,where $\alpha, \beta, \gamma \in N$,then $\alpha\beta\gamma$ is equal to $........$.

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 $y = y_1(x)$ and $y = y_2(x)$ be the solution curves of the differential equation $\frac{dy}{dx} = y + 7$ with initial conditions $y_1(0) = 0$ and $y_2(0) = 1$ respectively. Then the curves $y = y_1(x)$ and $y = y_2(x)$ intersect at

If the solution curve of the differential equation $(\tan^{-1} y - x) dy = (1 + y^2) dx$ passes through the point $(1, 0)$,then the abscissa of the point on the curve whose ordinate is $\tan(1)$ is