JEE Main 2026 Apr 2 Shift 1, Mathematics Q17: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 2, Shift 1Mathematics

If the curve $y = f (x)$ passes through the point $(1, e)$ and satisfies the differential equation $d y = y (2 + lo g_{e} x) d x$ , $x > 0$ , then $f (e)$ is equal to :

A
$e^{e}$
B
$e^{e^{2}}$
C
$e^{2 e}$
D
$e^{2^{e}}$

Solution

Separate variables: $\frac{d y}{y} = (2 + ln x) d x$ .

Integrate: $\int (2 + ln x) d x = 2 x + (x ln x - x) = x + x ln x$ .

So $ln y = x + x ln x + C$ .

Apply $y (1) = e$ : $1 = 1 + 0 + C \Rightarrow C = 0$ .

Hence $ln y = x (1 + ln x)$ , i.e. $y = e^{x + x l n x}$ .

$f (e) = e^{e + e \cdot 1} = e^{2 e}$ .

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JEE Main 2026 Apr 2 Shift 1, Mathematics Q17: Methods of Solving First Order, First Degree Differential Equation

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