JEE Main 2026 Jan 28 Shift 1, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Jan 28, Shift 1Mathematics

Let $y = y (x)$ be the solution of the differential equation

$x \frac{d y}{d x} - sin 2 y = x^{3} (2 - x^{3}) cos^{2} y, x \neq = 0$ .

If $y (2) = 0$ , then $tan (y (1))$ is equal to

Solution

Divide through by $x cos^{2} y$ :

$sec^{2} y \frac{d y}{d x} - 2 tan y \cdot \frac{1}{x} = x^{2} (2 - x^{3})$ (using $sin 2 y = 2 sin y cos y$ , so $sin 2 y / cos^{2} y = 2 tan y$ ).

Let $t = tan y$ , so $\frac{d t}{d x} = sec^{2} y \frac{d y}{d x}$ . The equation becomes linear in $t$ :

$\frac{d t}{d x} - \frac{2 t}{x} = x^{2} (2 - x^{3})$ .

Integrating factor: $IF = e^{\int - 2/ x d x} = e^{- 2 l n x} = \frac{1}{x ^{2}}$ .

$\frac{t}{x ^{2}} = \int \frac{1}{x ^{2}} \cdot x^{2} (2 - x^{3}) d x = \int (2 - x^{3}) d x = 2 x - \frac{x ^{4}}{4} + C$ .

Apply $y (2) = 0 \Rightarrow t = 0$ at $x = 2$ : $0 = 4 - 4 + C \Rightarrow C = 0$ .

$∴ tan y = 2 x^{3} - \frac{x ^{6}}{4}$ .

At $x = 1$ : $tan y (1) = 2 - \frac{1}{4} = \frac{7}{4}$ .

Option (2).

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