The differential equation $\frac{dy}{dx} = \frac{x+y}{1+x^2}$ is a . . . . . . differential equation.

The general solution of the differential equation,$y' + y\phi'(x) - \phi(x)\phi'(x) = 0$,where $\phi(x)$ is a known function,is: (where $c$ is an arbitrary constant)

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

If for $x \geq 0$,$y=y(x)$ is the solution of the differential equation $(x+1) dy = ((x+1)^{2} + y - 3) dx$ with $y(2) = 0$,then $y(3)$ 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:

The general solution of the differential equation $\frac{dy}{dx} + \left(\frac{3x^2}{1+x^3}\right)y = \frac{1}{x^3+1}$ is