Q.General solution of $\frac{dy}{dx}+y\tan x=\sec x$ is:
(A) $y\sec x=\tan x+c$
(B) $y\tan x=\sec x+c$
(C) $\tan x=y\tan x+c$
(D) $x\sec x=\tan y+c$

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$'s on one side and all the $x$'s on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

The Integrating Factor

The magic multiplier is

Why this one? Multiply the equation by $e^{\int P\,dx}$:

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$, because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$. So the equation becomes

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$:

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$, we read $P = \tfrac1x$, $Q = x$. Then $\int P\,dx = \log x$, so $\text{I.F.} = e^{\log x} = x$. The solution is

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