Maths Q17 – JEE-MAIN 2026

Home Maths Differential EquationsQuestion #171

If the curve

y = f (x)

passes through the point

(1, e)

and satisfies the differential equation

\frac{d y}{d x} = y (2 + \log_{e} x)

x > 0

, then

f (e)

is equal to :

Clue 1 of 3: Differential Equation Type

Recognize that the given differential equation is a first-order separable differential equation.

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● Video Walkthrough

Step 1: Separate variables and integrate✦ Active

The given differential equation is $\frac{d y}{d x} = y (2 + \log_{e} x)$ . Separate the variables:

\frac{d y}{y} = (2 + \log_{e} x) d x

Integrate both sides. For $\int \log_{e} x d x$ , use integration by parts ( $u = \log_{e} x, d v = d x ⟹ d u = \frac{1}{x} d x, v = x$ ), which gives $x \log_{e} x - x$ . Thus:

\int \frac{1}{y} d y = \int (2 + \log_{e} x) d x ⟹ \log_{e} | y | = 2 x + (x \log_{e} x - x) + C

Simplify the expression:

\log_{e} | y | = x + x \log_{e} x + C

Step 2: Apply initial condition to find the constant○ Expand

The curve passes through the point $(1, e)$ . Substitute $x = 1$ and $y = e$ into the general solution. Since $y = e > 0$ , $| y | = y$ .

\log_{e} e = 1 + 1 \log_{e} 1 + C

Since $\log_{e} e = 1$ and $\log_{e} 1 = 0$ :

1 = 1 + 0 + C ⟹ C = 0

Thus, the particular solution is:

\log_{e} y = x (1 + \log_{e} x) ⟹ y = e^{x (1 + \log_{e} x)}

Step 3: Evaluate f(e)○ Expand

Substitute $x = e$ into the particular solution to find $f (e)$ :

f (e) = e^{e (1 + \log_{e} e)}

Since $\log_{e} e = 1$ :

f (e) = e^{e (1 + 1)} = e^{e (2)} = e^{2 e}

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Differential Equation Type

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