Q.If the slope of the tangent to the curve at each of its points equals the sum of the abscissa and the product of the abscissa and ordinate, and the curve passes through $(0,1)$, find the equation of the curve.

Identify $P = \frac{1}{x}$ in the linear form $\frac{dy}{dx}+Py=Q$, then $\text{I.F.} = e^{\int P\,dx} = e^{\log x} = x$.

The given equation

$\frac{dy}{dx} + \frac{y}{x} = x^3 - 3$

is a linear differential equation of the standard form $\dfrac{dy}{dx} + Py = Q$, with

$P = \frac{1}{x}, \qquad Q = x^3 - 3.$

…

For $\dfrac{dy}{dx} + y = e^{-x}$ we have $P = 1$, so $\text{I.F.} = e^{\int 1\,dx} = e^{x}$.

A first-order linear differential equation has the standard form:

$\frac{dy}{dx} + Py = Q,$

where $P$ and $Q$ are functions of $x$ (or constants). Its integrating factor (I.F.) is $e^{\int P\,dx}$.

…

The equation is linear with $P(x) = \dfrac{1}{x}$, so its integrating factor is $e^{\int \frac{1}{x}\,dx} = e^{\log x} = x$, not $-x$. The statement is False.

The given equation $\dfrac{dy}{dx} + \dfrac{y}{x} = x^3$ is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with $P(x) = \dfrac{1}{x}$.

The integrating factor (I.F.) is …