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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Differential Equations

If the curve y = f(x) passes through the point (1, e) and satisfies...

If the curve y=f(x)y = f(x) passes through the point (1,e)(1, e) and satisfies the differential equation dy=y(2+log⁡ex) dxdy = y(2 + \log_e x)\,dx, x>0x > 0, then f(e)f(e) is equal to :

Options

  1. Aeee^e
  2. Bee2e^{e^2}
  3. Ce2ee^{2e}
  4. De2ee^{2e}

Correct answer

C. e2ee^{2e}

Step-by-step solution

Given differential equation is dyy=(2+log⁡ex)dx\dfrac{dy}{y} = (2 + \log_e x) dx

Integrating both sides, we get:

∫dyy=∫(2+log⁡ex)dx\int \dfrac{dy}{y} = \int (2 + \log_e x) dx

log⁡ey=2x+xlog⁡ex−x+C\log_e y = 2x + x \log_e x - x + C

log⁡ey=x+xlog⁡ex+C\log_e y = x + x \log_e x + C

Since the curve passes through (1,e)(1, e), substitute x=1x = 1 and y=ey = e:

log⁡ee=1+1⋅log⁡e1+C\log_e e = 1 + 1 \cdot \log_e 1 + C

1=1+0+C⇒C=01 = 1 + 0 + C \Rightarrow C = 0

The equation of the curve is log⁡ey=x+xlog⁡ex\log_e y = x + x \log_e x

To find f(e)f(e), substitute x=ex = e:

log⁡ey=e+elog⁡ee\log_e y = e + e \log_e e

log⁡ey=e+e(1)=2e\log_e y = e + e(1) = 2e

y=e2ey = e^{2e}

Answer: e2ee^{2e}
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