Differential Equations – Linear DE – JEE Main 29 Jan 2025 Shift 2

Solution:

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \tan x$.

The integrating factor (I.F.) is:
$$ \text{I.F.} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x $$

Multiplying the equation by the I.F. and integrating:
$$ y \sec x = \int \frac{2+\sec x}{(1+2\sec x)^2} \sec x \, dx $$
$$ y \sec x = \int \frac{2\sec x + \sec^2 x}{(1+2\sec x)^2} \, dx $$

Let’s simplify the integrand by converting to cosine:
$$ \frac{2\sec x + \sec^2 x}{(1+2\sec x)^2} = \frac{\frac{2}{\cos x} + \frac{1}{\cos^2 x}}{(1+\frac{2}{\cos x})^2} = \frac{\frac{2\cos x + 1}{\cos^2 x}}{\frac{(\cos x + 2)^2}{\cos^2 x}} = \frac{2\cos x + 1}{(\cos x + 2)^2} $$

Now we substitute $\cos x = \frac{1-t^2}{1+t^2}$ where $t = \tan(x/2)$. Then $dx = \frac{2dt}{1+t^2}$.
$$ \text{Integrand} = \frac{2(\frac{1-t^2}{1+t^2}) + 1}{(\frac{1-t^2}{1+t^2} + 2)^2} \cdot \frac{2}{1+t^2} $$
$$ = \frac{\frac{2-2t^2+1+t^2}{1+t^2}}{(\frac{1-t^2+2+2t^2}{1+t^2})^2} \cdot \frac{2}{1+t^2} = \frac{3-t^2}{(1+t^2)} \cdot \frac{(1+t^2)^2}{(t^2+3)^2} \cdot \frac{2}{1+t^2} $$
$$ = \frac{2(3-t^2)}{(t^2+3)^2} $$

To integrate $\int \frac{2(3-t^2)}{(t^2+3)^2} dt$, divide numerator and denominator by $t^2$:
$$ = 2 \int \frac{\frac{3}{t^2}-1}{(t+\frac{3}{t})^2} dt = -2 \int \frac{1-\frac{3}{t^2}}{(t+\frac{3}{t})^2} dt $$
Let $u = t + \frac{3}{t}$, then $du = (1 – \frac{3}{t^2}) dt$.
$$ = -2 \int \frac{du}{u^2} = -2\left(-\frac{1}{u}\right) = \frac{2}{u} = \frac{2}{t + \frac{3}{t}} $$

So, the solution is:
$$ y \sec x = \frac{2t}{t^2+3} + C $$
Given $f(\pi/3) = \frac{\sqrt{3}}{10}$. At $x = \pi/3$, $t = \tan(\pi/6) = \frac{1}{\sqrt{3}}$.
$$ \frac{\sqrt{3}}{10} \cdot \sec(\pi/3) = \frac{2(1/\sqrt{3})}{1/3 + 3} + C $$
$$ \frac{\sqrt{3}}{10} \cdot 2 = \frac{2/\sqrt{3}}{10/3} + C \Rightarrow \frac{\sqrt{3}}{5} = \frac{6}{10\sqrt{3}} + C \Rightarrow \frac{\sqrt{3}}{5} = \frac{\sqrt{3}}{5} + C $$
$$ C = 0 $$

For $f(\pi/4)$, we have $x = \pi/4$, so $t = \tan(\pi/8) = \sqrt{2}-1$.
$$ y \sec(\pi/4) = \frac{2(\sqrt{2}-1)}{(\sqrt{2}-1)^2 + 3} $$
$$ y\sqrt{2} = \frac{2(\sqrt{2}-1)}{(2+1-2\sqrt{2}) + 3} = \frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} $$
$$ y\sqrt{2} = \frac{\sqrt{2}-1}{3-\sqrt{2}} $$
$$ y = \frac{\sqrt{2}-1}{\sqrt{2}(3-\sqrt{2})} = \frac{\sqrt{2}-1}{3\sqrt{2}-2} $$
Rationalizing the denominator:
$$ y = \frac{(\sqrt{2}-1)(3\sqrt{2}+2)}{(3\sqrt{2}-2)(3\sqrt{2}+2)} = \frac{6 + 2\sqrt{2} – 3\sqrt{2} – 2}{18 – 4} = \frac{4 – \sqrt{2}}{14} $$

Ans. (4)