Q.Solve the differential equation: $\dfrac{\mathrm{d}y}{\mathrm{d}x} + y\sec x = \tan x$ OR Solve the differential equation: $(x+y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1$

Part 1: linear DE in $y$, integrating factor $\sec x+\tan x$. Part 2 (OR): rewrite as a linear DE in $x$ (treating $y$ as independent variable).

Part 1 — Solve $\dfrac{dy}{dx}+y\sec x=\tan x$:

This is linear in $y$: $\dfrac{dy}{dx}+Py=Q$ with $P=\sec x$, $Q=\tan x$.

Integrating factor: $\mathrm{IF} = e^{\int\sec x\,dx} = e^{\log|\sec x+\tan x|} = \sec x+\tan x$

General solution: $y\cdot\mathrm{IF} = \displaystyle\int Q\cdot\mathrm{IF}\,dx + c$

$y(\sec x+\tan x) = \int \tan x(\sec x+\tan x)\,dx = \int(\sec x\tan x + \tan^2x)\,dx$

$= \int\sec x\tan x\,dx + \int(\sec^2x-1)\,dx = \sec x + \tan x - x + c$

$\boxed{y(\sec x+\tan x) = \sec x+\tan x - x + c}$

Part 2 (OR) — Solve $(x+y)\dfrac{dy}{dx}=1$:

Rewrite as $\dfrac{dx}{dy} = x+y$, i.e. $\dfrac{dx}{dy} - x = y$ — linear in $x$ (treating $y$ as the independent variable), with $P=-1$, $Q=y$.

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