JEE Main 2025 Apr 4 Shift 2, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Apr 4, Shift 2Mathematics

If a curve $y = y (x)$ passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} cot y - e^{x} csc y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at $x = 2$ , the value of $cos y$ is:

A
$\frac{2 e ^{2} - e}{64}$
B
$\frac{2 e ^{2} + e}{64}$
C
$\frac{2 e ^{2} - e}{128}$
D
$\frac{2 e ^{2} + e}{128}$

Solution

Rewrite the equation as $\frac{d y}{d x} = \frac{7 cot y}{x} - \frac{e ^{x} csc y}{x ^{5}}$ , i.e.

$sin y \frac{d y}{d x} - \frac{7 cos y}{x} = - \frac{e ^{x}}{x ^{5}}$ .

Let $t = - cos y$ , so $\frac{d t}{d x} = sin y \frac{d y}{d x}$ . The equation becomes the linear ODE

$\frac{d t}{d x} + \frac{7 t}{x} = - \frac{e ^{x}}{x ^{5}}$ .

Integrating factor: $μ = x^{7}$ . Then

$\frac{d}{d x} (t x^{7}) = - x^{2} e^{x} \Rightarrow t x^{7} = - \int x^{2} e^{x} d x = - (x^{2} - 2 x + 2) e^{x} + C$ .

So $cos y \cdot x^{7} = (x^{2} - 2 x + 2) e^{x} - C$ . At $x = 1, y = π /2$ : $cos y = 0$ , giving $C = e$ .

At $x = 2$ : $cos y \cdot 128 = (4 - 4 + 2) e^{2} - e = 2 e^{2} - e$ .

$∴ cos y = \frac{2 e ^{2} - e}{128}$ .

Concept behind this question

Methods of Solving First Order, First Degree Differential EquationNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 4 Shift 2, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

Concept behind this question

More previous year questions on Methods of Solving First Order, First Degree Differential Equation

Practice more Mathematics