Q.Solve the following differential equation: The Integrating Factor of the differential equation $x \frac{dy}{dx} - y = 2x^2$ is (A) $e^{-x}$ (B) $e^{-y}$ (C) $\frac{1}{x}$ (D) $x$

Concept: Integrating Factor Method — For a linear first-order ODE of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$.

First, rewrite the given equation in standard form:

Here $P(x) = -\frac{1}{x}$.

Compute the integrating factor: …

The Integrating Factor method is the go-to tool for any first-order linear differential equation — that is, any equation that can be written in the form $\frac{dy}{dx} + P(x) y = Q(x)$. The idea is simple: we want to multiply the entire equation by some function $\mu(x)$ so that the left-hand side becomes the exact derivative of $\mu(x) y$. Once that happens, we can integrate both sides directly.

Why does this work? Because if you choose $\mu(x) = e^{\int P(x) dx}$, then by the product rule:

which is exactly $\mu(x)$ times the left-hand side of the standard form. So the integrating factor turns a tricky sum of derivatives into a single, clean derivative.

Now let's apply this to the given equation.

1. Rewrite in standard form. The equation is $x \frac{dy}{dx} - y = 2x^2$. Divide through by $x$ (assuming $x \neq 0$):

Here $P(x) = -\frac{1}{x}$ and $Q(x) = 2x$.

2. Compute the integrating factor. The formula is $\mu(x) = e^{\int P(x) dx}$. So:

We only need one integrating factor (the constant can be ignored), so take:

Since we usually work with positive $x$ or take the absolute value as understood, the standard integrating factor is $\frac{1}{x}$. …

Method: Finding the integrating factor of a linear equation

For "the integrating factor is ..." questions, the only work is to reach standard form and compute $e^{\int P\,dx}$.

Steps

Step 1: Divide to make the coefficient of $\dfrac{dy}{dx}$ equal to $1$.

This is the step most easily skipped — $x\dfrac{dy}{dx}-y=2x^2$ must first become $\dfrac{dy}{dx}-\dfrac{1}{x}y=2x$.

Step 2: Read off $P(x)$.

Here $P=-\dfrac{1}{x}$. …

Common Mistakes

Mistake 1: Reading $P=-1$ straight from $xy'-y=2x^2$.

Why it's wrong: without dividing by $x$ you wrongly get I.F. $=e^{-x}$ (the tempting option A). Correct approach: divide by $x$ first to get $P=-\tfrac1x$.

Mistake 2: Sign error in $e^{-\log|x|}$. …