Differential EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Differential Equation Through (1,e): f(e) = e^(2e) | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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
Aeee^{e}
Bee2e^{e^2}
Ce2ee^{2e}correct
De2ee^{2e}
Solution
Step 1: Separate variables:
dyy=(2+log⁡ex) dx.\frac{dy}{y}=(2+\log_e x)\,dx.
 Step 2: Integrate. Note ∫log⁡ex dx=xlog⁡ex−x\displaystyle\int\log_e x\,dx=x\log_e x-x:
log⁡ey=2x+(xlog⁡ex−x)+C=x+xlog⁡ex+C.\log_e y=2x+(x\log_e x-x)+C=x+x\log_e x+C.
 Step 3: Apply (1,e)(1,e): log⁡ee=1\log_e e=1, so 1=1+1⋅0+C⇒C=01=1+1\cdot0+C\Rightarrow C=0. Hence
log⁡ey=x+xlog⁡ex ⇒ y=e x+xlog⁡ex.\log_e y=x+x\log_e x\ \Rightarrow\ y=e^{\,x+x\log_e x}.
 Step 4: Evaluate at x=ex=e (using log⁡ee=1\log_e e=1):
f(e)=e e+e⋅1=e e+e=e2e.f(e)=e^{\,e+e\cdot1}=e^{\,e+e}=e^{2e}.
 Correct answer: (3)
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