If a curve y=y(x) passes through the point (1, (π)/(2)) and satisfies the differential equation (7 x^4 y-e^x cosec y) (d x)/(d y)=x^5, x ≥ 1 , then at x=2 , the value of y is:

Correct answer: C. 2 e^2-e 128 aligned & (d y)/(d x)=(7 y)/(x)- e^x cosec y x^5 & (d y)/(d x)=(7 y)/( y · x)-(e^x)/( y x^5) & y (d y)/(d x)- y · (7)/(x)=(-e^x)/(x^5) & let - y=t & y (d y)/(d x)=(d t)/(d x) aligned aligned & (d t)/(d x)+(7 t)/(x)=(-e^x)/(x^5) & I.F. =x^7 & t. x^7=-∫ x^2 e^x d x & y x^7=x^2 e^x-2 ∫ x e^x d x & y x^7=x^2 e^x-2 x e^x+2 e^x+c & x=1, y=(π)/(2), c=-e & y=(2 e^2-e)/(128) & option (3) aligned

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Differential Equations

If a curve y=y(x) passes through the point (1, (π)/(2)) and...

If a curve

y=y(x)

passes through the point \left(1, \frac{\pi}{2}\right)

and satisfies the differential equation \left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1

, then at

x=2

, the value of \cos y is:

Options

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

Correct answer

C. $\frac{2 \mathrm{e}^2-\mathrm{e}}{128}$

Step-by-step solution

\begin{aligned} & \frac{d y}{d x}=\frac{7 \cot y}{x}-\frac{e^x \operatorname{cosec} y}{x^5} \\ & \frac{d y}{d x}=\frac{7 \cot y}{\sin y \cdot x}-\frac{e^x}{\sin y x^5} \\ & \sin y \frac{d y}{d x}-\cos y \cdot \frac{7}{x}=\frac{-e^x}{x^5} \\ & \text{ let }-\cos y=t \\ & \sin y \frac{d y}{d x}=\frac{d t}{d x}\end{aligned}

\begin{aligned} & \frac{d t}{d x}+\frac{7 t}{x}=\frac{-e^x}{x^5} \\ & \text{ I.F. }=x^7 \\ & \text{ t. } x^7=-\int x^2 e^x d x \\ & \operatorname{\cos y} x^7=x^2 e^x-2 \int x e^x d x \\ & \operatorname{\cos y} x^7=x^2 e^x-2 x e^x+2 e^x+c \\ & x=1, y=\frac{\pi}{2}, c=-e \\ & \operatorname{\cos y}=\frac{2 e^2-e}{128} \\ & \text{ option }(3)\end{aligned}

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