The solution of $\frac{dy}{dx} + p(x)y = 0$ is

The general solution of the differential equation $(2x - 10y^3) dy + y dx = 0, y \neq 0$ is

Let $\alpha x = \exp(x^\beta y^\gamma)$ be the solution of the differential equation $2x^2 y \frac{dy}{dx} - (1 - xy^2) = 0$,for $x > 0$ and $y(2) = \sqrt{\log_e 2}$. Then $\alpha + \beta - \gamma$ equals:

Suppose $f(x)$ is a differentiable real function such that $f(x) + f'(x) \le 1$ for all $x$ and $f(0)=0$. The largest possible value of $f(1)$ is

Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ is equal to:

If the length of the sub-tangent at any point $P(x, y)$ on a curve $f(x, y) = 0$ is $x + 7y^2$,then $f(x, y) =$