Q.Solve: $\dfrac{dy}{dx}+\dfrac{y}{x}=x^3-3$

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

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

The integrating factor is

$\text{I.F.} = e^{\int P\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\log x} = x.$

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 $e^{\int P\,dx}$.

Comparing $\dfrac{dy}{dx} + y = e^{-x}$ with the standard form, the coefficient of $y$ is $P = 1$ (and $Q = e^{-x}$).

Compute the integrating factor:

$\text{I.F.} = e^{\int P\,dx} = e^{\int 1\,dx} = e^{x}.$

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

$\text{I.F.} = e^{\int P\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\log|x|} = x.$

So the correct integrating factor is $x$, whereas the statement claims $-x$.