Differential Equations – Linear Differential Equations – JEE Main 28 Jan 2026 Shift 1

Solution:

Divide the differential equation by $x \cos^2 y$:
$$\frac{1}{\cos^2 y} \frac{dy}{dx} – \frac{\sin 2y}{x \cos^2 y} = x^2(2-x^3)$$
$$\sec^2 y \frac{dy}{dx} – \frac{2 \sin y \cos y}{x \cos^2 y} = 2x^2 – x^5$$
$$\sec^2 y \frac{dy}{dx} – \frac{2 \tan y}{x} = 2x^2 – x^5$$

Let $\tan y = t$. Then $\sec^2 y \frac{dy}{dx} = \frac{dt}{dx}$.
Substitute into the equation:
$$\frac{dt}{dx} – \frac{2}{x}t = 2x^2 – x^5$$
This is a Linear Differential Equation (LDE) of the form $\frac{dt}{dx} + P(x)t = Q(x)$.
Integrating Factor (I.F.):
$$I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = e^{\ln x^{-2}} = \frac{1}{x^2}$$

Solution of LDE:
$$t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$
$$t \cdot \frac{1}{x^2} = \int (2x^2 – x^5) \frac{1}{x^2} dx + C$$
$$\frac{t}{x^2} = \int (2 – x^3) dx + C$$
$$\frac{\tan y}{x^2} = 2x – \frac{x^4}{4} + C$$
$$\tan y = 2x^3 – \frac{x^6}{4} + Cx^2$$

Given $y(2) = 0$. Substitute $x=2, y=0$:
$$\tan(0) = 2(8) – \frac{64}{4} + C(4)$$
$$0 = 16 – 16 + 4C \Rightarrow C = 0$$
So the particular solution is:
$$\tan y = 2x^3 – \frac{x^6}{4}$$

Find $\tan(y(1))$ by substituting $x=1$:
$$\tan(y(1)) = 2(1)^3 – \frac{(1)^6}{4}$$
$$\tan(y(1)) = 2 – \frac{1}{4} = \frac{7}{4}$$

Ans. (2)