JEE Mathematics 2023 Question with Solution

Given: $\frac{dy}{dx} + y \tan x = x \sec x$ with $y(0)=1$.

Find: $y\left(\frac{\pi}{6}\right)$.

For the linear differential equation, the integrating factor is

$$I.F.=\sec x$$

Multiplying throughout by $\sec x$,

$$\frac{d}{dx}(y\sec x)=x\sec^2 x$$

Integrating,

$$y\sec x = x\tan x - \ln(\sec x) + C$$

Using the initial condition $y(0)=1$,

$$1 = 0 - \ln(1) + C \Rightarrow C=1$$

Hence,

$$y\sec x = x\tan x - \ln(\sec x) + 1$$

Now at $x=\frac{\pi}{6}$,

$$\sec \frac{\pi}{6}=\frac{2}{\sqrt{3}}, \qquad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$$

So,

$$y\cdot \frac{2}{\sqrt{3}} = \frac{\pi}{6\sqrt{3}} - \ln\left(\frac{2}{\sqrt{3}}\right) + 1$$

Therefore,

$$y = \frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\left(\frac{2}{e\sqrt{3}}\right)$$

the solution concludes that the correct option is A. Note that the simplified value written in the solution differs in sign form from the printed option text, but the source marks A as correct.

Since

$$\frac{dy}{dx}+y\tan x = x\sec x$$

we choose integrating factor

$$\exp\left(\int \tan x \, dx\right)=\exp\left(\ln \sec x\right)=\sec x$$

Then

$$\sec x\frac{dy}{dx} + y\sec x\tan x = x\sec^2 x$$

The left side is

$$\frac{d}{dx}(y\sec x)$$

Hence,

$$\frac{d}{dx}(y\sec x)=x\sec^2 x$$

Now integrate by parts for $\int x\sec^2 x \, dx$:

$$\int x\sec^2 x \, dx = x\tan x - \int \tan x \, dx = x\tan x - \ln(\sec x)$$

So,

$$y\sec x = x\tan x - \ln(\sec x) + C$$

Using $x=0$ and $y=1$,

$$1 = C$$

Thus,

$$y\sec x = x\tan x - \ln(\sec x) + 1$$

Substitute $x=\frac{\pi}{6}$ and divide by $\sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}$ to get

$$y\left(\frac{\pi}{6}\right)=\frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\left(\frac{2}{e\sqrt{3}}\right)$$

Therefore, the correct option is A according to the solution.