Differential Equations – Linear Differential Equation – JEE Main 21 Jan 2026 Shift 2

Solution:

Rewrite the differential equation in linear form $\frac{dy}{dx} + Py = Q$:
Multiply by $\cos x$:
$$\frac{dy}{dx} – 2y \cos x = 2 \cos x + 3 \sin x \cos x$$

Find the Integrating Factor (I.F.):
$$I.F. = e^{\int -2 \cos x dx} = e^{-2 \sin x}$$

The general solution is given by:
$$y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$$

$$y \cdot e^{-2 \sin x} = \int (2 \cos x + 3 \sin x \cos x) e^{-2 \sin x} dx$$

Let $\sin x = t$, then $\cos x dx = dt$.
$$y \cdot e^{-2 \sin x} = \int (2 + 3t) e^{-2t} dt$$

Using integration by parts

$$\int (2 + 3t) e^{-2t} dt = (2+3t)\left(\frac{e^{-2t}}{-2}\right) – \int \left(\frac{e^{-2t}}{-2}\right) 3 dt$$

$$= -\frac{1}{2}(2+3t)e^{-2t} + \frac{3}{2} \int e^{-2t} dt$$

$$= -\frac{1}{2}(2+3t)e^{-2t} + \frac{3}{2} \left(\frac{e^{-2t}}{-2}\right)$$

$$= -e^{-2t} \left[ \frac{2+3t}{2} + \frac{3}{4} \right]$$

$$= -e^{-2t} \left[ \frac{4+6t+3}{4} \right] = -\frac{7+6t}{4} e^{-2t}$$

Substitute $t = \sin x$ back:
$$y \cdot e^{-2 \sin x} = -\frac{7+6\sin x}{4} e^{-2 \sin x} + C$$

$$y = -\frac{7+6\sin x}{4} + C e^{2 \sin x}$$

Apply the initial condition $y(0) = -\frac{7}{4}$:
$$-\frac{7}{4} = -\frac{7+6(0)}{4} + C e^{2(0)}$$

$$-\frac{7}{4} = -\frac{7}{4} + C \Rightarrow C = 0$$

So, the particular solution is:
$$y(x) = -\frac{7+6\sin x}{4}$$

Now find $y(\frac{\pi}{6})$:
$$y\left(\frac{\pi}{6}\right) = -\frac{7+6\sin(\pi/6)}{4}$$

$$y\left(\frac{\pi}{6}\right) = -\frac{7+6(1/2)}{4} = -\frac{7+3}{4} = -\frac{10}{4} = -\frac{5}{2}$$

Ans. (1)