Q.Solve the following differential equation: $\cos^2 x \frac{dy}{dx} + y = \tan x \quad \left(0 \le x < \frac{\pi}{2}\right)$

First-order linear equation; no initial condition is given, so we find the general solution.

Standard form: divide by $\cos^2 x$:

Integrating factor: $\mu = e^{\int\sec^2 x\,dx} = e^{\tan x}$.

Integrate: $\frac{d}{dx}\left(y\,e^{\tan x}\right) = \tan x\sec^2 x\,e^{\tan x}$. With $t=\tan x$, $\int t\,e^{t}\,dt = e^{t}(t-1)$, so …

Spotting the type

After dividing by $\cos^2 x$, the equation has the form $\frac{dy}{dx}+P(x)y=Q(x)$ — a first-order linear equation, solved with an integrating factor.

Standard form

Since $\frac{1}{\cos^2 x}=\sec^2 x$,

so $P=\sec^2 x$ and $Q=\tan x\sec^2 x$.

Integrating factor

Multiply and integrate

Multiplying makes the left side an exact derivative:

$\frac{d}{dx}\left(y\,e^{\tan x}\right) = \tan x\sec^2 x\,e^{\tan x}.$ …

Method: Integrating factor after dividing by a leading coefficient

When the coefficient of $\dfrac{dy}{dx}$ is a function (like $\cos^2 x$), divide first, then apply the integrating factor.

Steps

Step 1: Divide to reach standard form.

Divide the whole equation by the coefficient of $\dfrac{dy}{dx}$ so it becomes $\dfrac{dy}{dx}+P\,y=Q$. Simplify with identities (e.g. $\tfrac{1}{\cos^2x}=\sec^2x$).

Step 2: Integrating factor.

For $P=\sec^2x$ this is $e^{\tan x}$.

Step 3: Multiply and integrate. …

Common Mistakes

Mistake 1: Forgetting to divide by $\cos^2 x$ first.

Why it's wrong: the coefficient of $\dfrac{dy}{dx}$ must be $1$ before reading $P$; skipping this gives the wrong $P$ and I.F. Correct approach: divide through, using $\tfrac{1}{\cos^2x}=\sec^2x$, to get $P=\sec^2x$.

Mistake 2: Struggling with $\int\tan x\sec^2x\,e^{\tan x}\,dx$. …