Let g be a differentiable function such that integral from 0 to x of g(t) dt = x - integral from 0 to x of tg(t) dt, x >= 0 and let y = y(x) satisfy the differential equation dy/dx - y tan x = 2(x+1)sec x g(x), x in [0, pi/2). If y(0) = 0, then y(pi/3) is equal to

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2025

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Home Maths Differential EquationsQuestion #184

Let g be a differentiable function such that

\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t, x \geq 0

and let

y = y (x)

satisfy the differential equation

\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x), x \in [0, \frac{π}{2})

. If

y (0) = 0

, then

y (\frac{π}{3})

is equal to

Clue 1 of 3: Leibniz Rule for Differentiation

Differentiate the given integral equation with respect to $x$ to determine the function $g (x)$ .

🧠 Full Solution Path

Step 1: Determine the function

g (x)

✦ Active

Differentiate the given integral equation $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t$ with respect to $x$ using Leibniz rule.

g (x) = 1 - x g (x)

g (x) (1 + x) = 1 ⟹ g (x) = \frac{1}{1 + x}

Step 2: Solve the differential equation for

y (x)

○ Expand

Substitute $g (x) = \frac{1}{1 + x}$ into the differential equation $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x)$ . This is a linear first-order ODE. Calculate the integrating factor (IF).

\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x (\frac{1}{1 + x})

\frac{d y}{d x} - y \tan x = 2 \sec x

P (x) = - \tan x, Q (x) = 2 \sec x

I F = e^{\int - \tan x d x} = e^{\ln | \cos x |} = \cos x for x \in [0, \frac{π}{2})

y \cdot I F = \int Q (x) \cdot I F d x + C

y \cos x = \int (2 \sec x) (\cos x) d x + C

y \cos x = \int 2 d x + C ⟹ y \cos x = 2 x + C

💡 Teacher's Secret Hint

Remember to consider the domain of $x$ when simplifying the absolute value in the integrating factor.

Step 3: Apply initial condition and evaluate

y (\frac{π}{3})

○ Expand

Use the initial condition $y (0) = 0$ to find the constant $C$ , then substitute $x = \frac{π}{3}$ into the particular solution for $y (x)$ .

y (0) = 0 ⟹ 0 \cdot \cos (0) = 2 (0) + C ⟹ C = 0

y (x) = \frac{2 x}{\cos x} = 2 x \sec x

y (\frac{π}{3}) = 2 (\frac{π}{3}) \sec (\frac{π}{3}) = 2 (\frac{π}{3}) (2) = \frac{4 π}{3}

💡 Teacher's Secret Hint

Ensure correct evaluation of trigonometric functions at standard angles.

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Maths Question 20 – JEE-MAIN 2025

Clue 1 of 3: Leibniz Rule for Differentiation

🧠 Full Solution Path