The general solution of the differential equation $\frac{dx}{dy} + \frac{x}{y} = x^2$ is

Integrating factor of $(x+2 y^3) \frac{d y}{d x}=y^2$ is

Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. If $y(0)=0$,then $y(2)$ is equal to

If $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$,then $\frac{d^2y}{dx^2} - 2m \frac{dy}{dx} + m^2y$ is equal to:

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:

Let $f: R \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in R$ and $f(0) \ne 0$. Let $g: [1, \infty) \to R$ be a differentiable function such that $x^2 g(x) = \int_1^x (t^2 f(t) - t g(t)) dt$. Then $g(2)$ is equal to: