Let a curve y=f(x) pass through the points (0,5) and ( _e 2, k) . If the curve satisfies the differential equation 2(3+y) e^ 2 x d x-(7+e^ 2 x ) d y=0 , then k is equal to

Correct answer: C. 8 aligned & (d y)/(d x)= 2(3+y) · e^ 2 x 7+e^ 2 x & (d y)/(d x)- 2 y e^ 2 x 7+e^ 2 x = 6 · e^ 2 x 7+e^ 2 x & I.F. =e^ -∫ 2 e^ 2 x 7+e^ 2 x d x & e^ - (7+e^ 2 x ) & = 1 7+e^ 2 x & y · 1 7+e^ 2 x =∫ 6 e^ 2 x (7+e^ 2 x )^2 d x & y 7+e^ 2 x = -3 7+e^ 2 x +C & y(0)=5 & (5)/(8)=(-3)/(8)+C & C=1 & y=-3+7+e^ 2 x & y=e^ 2 x +4 & k=8 aligned

JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Differential Equations

Let a curve y=f(x) pass through the points (0,5) and ( _e 2, k) ....

Let a curve

y=f(x)

pass through the points

(0,5) and \left(\log _e 2, k\right)

. If the curve satisfies the differential equation

2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0

, then

k

is equal to

Options

A $4$
B $32$
C $8$
D $16$

Correct answer

C. $8$

Step-by-step solution

\begin{aligned} & \frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}} \\ & \frac{d y}{d x}-\frac{2 y e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}} \\ & \text{ I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}} d x} \\ & \Rightarrow e^{-\ln \left(7+e^{2 x}\right)} \\ & =\frac{1}{7+e^{2 x}} \\ & y \cdot \frac{1}{7+e^{2 x}}=\int \frac{6 e^{2 x}}{\left(7+e^{2 x}\right)^2} d x \\ & \frac{y}{7+e^{2 x}}=\frac{-3}{7+e^{2 x}+C} \\ & \therefore y(0)=5 \\ & \Rightarrow \frac{5}{8}=\frac{-3}{8}+C \\ & \Rightarrow C=1 \\ & \therefore y=-3+7+e^{2 x} \\ & y=e^{2 x}+4 \\ & \therefore k=8\end{aligned}

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