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

The equation is linear in $y$ with $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \sec x$ and $Q(x) = \tan x$.

Step 1: Find the integrating factor

Step 2: Multiply through

The left side is $\frac{d}{dx}\big[(\sec x + \tan x) y\big]$.

Step 3: Integrate both sides

Compute: $\sec x \tan x + \tan^2 x = \sec x \tan x + (\sec^2 x - 1)$. …

The equation $\frac{dy}{dx} + (\sec x) y = \tan x$ is a classic first-order linear differential equation. The standard form is $\frac{dy}{dx} + P(x) y = Q(x)$, and the method of integrating factors is designed exactly for this.

Why does the integrating factor work? The idea is to multiply the entire equation by a cleverly chosen function $I(x)$ so that the left-hand side becomes the derivative of a product — specifically, $\frac{d}{dx}[I(x) \cdot y]$. This turns the problem into a simple integration. The magic is that $I(x) = e^{\int P(x) \, dx}$ always does the job, because:

• $\frac{d}{dx}(I y) = I \frac{dy}{dx} + I' y$
• If $I' = I P$, then this equals $I \frac{dy}{dx} + I P y = I(\frac{dy}{dx} + P y)$
• So multiplying by $I$ collapses the left side into a single derivative.

Let's apply this step by step.

1. Identify $P(x)$ and $Q(x)$.

   Here, $P(x) = \sec x$ and $Q(x) = \tan x$. The domain is $0 \le x < \frac{\pi}{2}$, where both $\sec x$ and $\tan x$ are positive and well-defined.

2. Compute the integrating factor $I(x) = e^{\int P(x) \, dx}$.

   We need $\int \sec x \, dx$. This is a standard integral with a clever trick:

Notice the numerator is exactly the derivative of the denominator: $\frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x$. So:

Since $x$ is in the first quadrant, $\sec x + \tan x > 0$, so we can drop the absolute value. Thus:

3. Multiply the ODE by the integrating factor. The original equation is:

Multiply through by $I(x) = \sec x + \tan x$:

The left-hand side should now be $\frac{d}{dx}[(\sec x + \tan x) y]$. Let's verify quickly:

$\frac{d}{dx}[(\sec x + \tan x) y] = (\sec x + \tan x) \frac{dy}{dx} + (\sec x \tan x + \sec^2 x) y$

And $(\sec x + \tan x)(\sec x) = \sec^2 x + \sec x \tan x$, which matches. Perfect.

4. Rewrite and integrate. The equation becomes: $\frac{d}{dx}[(\sec x + \tan x) y] = (\sec x + \tan x) \tan x$ …

Method: Integrating factor when $P$ involves $\sec x$

Use this for a linear equation where computing $\int P\,dx$ needs a standard trig integral such as $\int\sec x\,dx$.

Steps

Step 1: Standard form; identify $P$ and $Q$.

Here the coefficient of $\dfrac{dy}{dx}$ is already $1$.

Step 2: Recall the awkward integral.

Step 3: Multiply and integrate the right side. …

Common Mistakes

Mistake 1: Not knowing $\int\sec x\,dx=\log|\sec x+\tan x|$.

Why it's wrong: without this the integrating factor $\sec x+\tan x$ cannot be found. Correct approach: memorise this standard integral; the I.F. is exactly $\sec x+\tan x$.

Mistake 2: Forgetting $\tan^2x=\sec^2x-1$ when integrating the right side.

Why it's wrong: $\int\tan^2x\,dx=\tan x-x$, not $\tan x$; missing the $-x$ term changes the answer. Correct approach: rewrite $\tan^2x$ before integrating. …