Differential Equations JEE 2025: Why the Key's 16 Fails

What is the differential equations JEE 2025 question, and is its answer verified?

The supplied differential equations JEE 2025 equation cannot give the answer key’s 16. The initial-value problem is solvable, but the answer and exam attribution remain unverified.

The supplied task defines a function satisfying:

f^{'} (x) + \frac{f (x)}{x - 1} = \frac{x}{\sqrt{1 - x^{2}}}, - 1 < x < 1, f (0) = 0 .

It asks for:

α^{2}, 6 \int_{- 1 / 2}^{1 / 2} f (x) d x = 2 π - α .

This is a numerical-answer question with no supplied answer options. 16 is the supplied key’s answer, not a mathematically established result.

The record labels the question 2025 but gives the date 2023-01-24. Do not treat the year, date or shift as verified until the original paper is checked. This is a worked source check, not a resolved, verified PYQ solution.

How do you find the integrating factor without losing its sign?

On the stated domain, the integrating factor simplifies to: $I F = 1 - x .$

Its negative also works, provided you multiply both sides consistently. The error is not choosing a negative integrating factor; it is changing the sign on only one side.

Compare the equation with the standard linear form:

y^{'} + P (x) y = Q (x), P (x) = \frac{1}{x - 1}, Q (x) = \frac{x}{\sqrt{1 - x^{2}}} .

Calculate the integrating factor:

I F = \exp (\int \frac{d x}{x - 1}) = \exp (\ln | x - 1 |) = | x - 1 | .

Use the given domain:

- 1 < x < 1 ⟹ | x - 1 | = 1 - x .

Multiplication gives:

(1 - x) y^{'} - y = \frac{x (1 - x)}{\sqrt{1 - x^{2}}} .

The left side is a product derivative:

\frac{d}{d x} [(1 - x) y] = \frac{x (1 - x)}{\sqrt{1 - x^{2}}} .

An integrating factor is not unique. The alternative differs by a nonzero constant factor, so it is equally valid: $x - 1 = - (1 - x) .$

How do you integrate and apply the initial condition?

Integrate the product derivative, retain the constant, then use the initial condition to determine it. The first integral is direct; the second needs a sine substitution. Keep the pieces separate to avoid losing the minus sign before the second integral.

Split the right-hand integral:

(1 - x) f (x) = \int \frac{x}{\sqrt{1 - x^{2}}} d x - \int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x .

The first antiderivative is:

\int \frac{x}{\sqrt{1 - x^{2}}} d x = - \sqrt{1 - x^{2}} .

For the second, use:

x = \sin θ, d x = \cos θ d θ, - \frac{π}{2} < θ < \frac{π}{2} .

Cosine is positive on this interval, so:

\int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x = \int \sin^{2} θ d θ = \frac{θ}{2} - \frac{\sin 2 θ}{4} = \frac{\arcsin x - x \sqrt{1 - x^{2}}}{2} .

Combine the pieces with one arbitrary constant:

(1 - x) f (x) = - \sqrt{1 - x^{2}} - \frac{\arcsin x}{2} + \frac{x \sqrt{1 - x^{2}}}{2} + C .

Now apply the initial condition:

x = 0, f (0) = 0 ⟹ 0 = - 1 + C ⟹ C = 1 .

The solution is:

f (x) = \frac{1 - \sqrt{1 - x^{2}} - \frac{1}{2} \arcsin x + \frac{1}{2} x \sqrt{1 - x^{2}}}{1 - x} .

Check it immediately:

f (0) = \frac{1 - 1 - 0 + 0}{1} = 0 .

Differentiating the product expression recovers the original differential equation. The solution therefore satisfies both the equation and the initial condition.

Why does the definite integral rule out the answer 16?

The equation bounds the magnitude of the requested integral by approximately 0.8660. The key requires approximately 2.2832 or 10.2832, so a bound alone proves the conflict. An exact evaluation is unnecessary to reject 16.

Define:

I = \int_{- 1 / 2}^{1 / 2} f (x) d x .

Integrating the product derivative from the initial point gives:

(1 - x) f (x) = \int_{0}^{x} \frac{t (1 - t)}{\sqrt{1 - t^{2}}} d t .

Throughout the integration interval:

| t | \leq \frac{1}{2}, 1 - t \leq \frac{3}{2}, \sqrt{1 - t^{2}} \geq \frac{\sqrt{3}}{2} .

Therefore:

| \frac{t (1 - t)}{\sqrt{1 - t^{2}}} | \leq \sqrt{3} | t | .

For either sign of the endpoint, integrate this bound along the interval between zero and that endpoint:

| (1 - x) f (x) | \leq \frac{\sqrt{3} x^{2}}{2} .

The denominator is bounded away from zero:

1 - x \geq \frac{1}{2} ⟹ | f (x) | \leq \sqrt{3} x^{2} .

Hence:

| 6 I | \leq 6 \int_{- 1 / 2}^{1 / 2} | f (x) | d x \leq 6 \sqrt{3} \int_{- 1 / 2}^{1 / 2} x^{2} d x = 6 \sqrt{3} (\frac{1}{12}) = \frac{\sqrt{3}}{2} \approx 0.8660 .

Accepting the supplied key would require:

α^{2} = 16 ⟹ α = 4 or - 4 .

Thus:

6 I = 2 π - 4 \approx 2.2832 or 6 I = 2 π + 4 \approx 10.2832 .

Both exceed the proven upper bound. The supplied equation and supplied answer cannot both be correct.

Obtain the original paper or a corrected equation/key before treating this as an official solution. Publication as a verified PYQ solution must wait; guessing a repair would hide the conflict.

What happens if you drop the integration constant?

Dropping the constant produces a function that fails the initial condition immediately. An indefinite integral describes a family of solutions; the initial condition selects one member. Omitting that step also changes the requested definite integral.

If the constant is omitted, it is effectively set to zero:

f_{w r o n g} (0) = - 1,

instead of: $f (0) = 0 .$

Setting the constant to zero rather than its required value changes the entire function:

C = 0 instead of C = 1, f_{w r o n g} (x) - f (x) = - \frac{1}{1 - x} .

It changes the definite integral by:

I_{w r o n g} - I = - \int_{- 1 / 2}^{1 / 2} \frac{d x}{1 - x} = \ln \frac{1 / 2}{3 / 2} = - \ln 3 .

This is an incorrect route, not a listed wrong option: the question is numerical and has no supplied distractors. Substitute into the initial condition before starting any definite integration.

Which two related questions should you solve next?

Use these original related practice questions, not additional JEE PYQs, to test method selection. The first repeats the integrating-factor structure with easier integration. For the second, pick the homogeneous substitution because its right side depends only on the ratio of the dependent and independent variables.

How do you solve the related linear initial-value problem?

Use the same integrating factor, integrate the polynomial and apply the initial condition. This isolates the method from the trigonometric integration in the supplied problem.

Solve:

y^{'} + \frac{y}{x - 1} = x, - 1 < x < 1, y (0) = 0 .

The compact route is:

I F = 1 - x, [(1 - x) y]^{'} = x - x^{2} .

(1 - x) y = \frac{x^{2}}{2} - \frac{x^{3}}{3} + C, C = 0 .

Therefore:

y = \frac{x^{2} / 2 - x^{3} / 3}{1 - x} .

When should you use the homogeneous substitution instead?

Use it when the right side depends only on the ratio of the dependent and independent variables, as it does here. The substitution cancels that ratio and leaves a direct integral.

Solve:

y^{'} = 1 + \frac{y}{x}, x > 0, y (1) = 0 .

Substitute and simplify:

y = v x, y^{'} = v + x v^{'}, v + x v^{'} = 1 + v, x v^{'} = 1 .

Integrate and apply the condition:

v = \ln x + C, y = x (\ln x + C), y (1) = 0 ⟹ C = 0 .

y = x \ln x .

The main problem is linear in the dependent variable; this practice equation directly suits the homogeneous substitution. Redo both without looking, then verify each answer in its differential equation and initial condition.

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Frequently asked questions

Is 16 the correct answer to this differential equations JEE 2025 question?

No, not for the supplied equation and initial condition. Writing I for the requested integral, the equation gives |6I| ≤ sqrt(3)/2, approximately 0.8660. The answer α² = 16 would require 6I to be approximately 2.2832 or 10.2832, contradicting that bound.

Why is the integrating factor 1-x instead of x-1?

For the coefficient 1/(x-1), the exponential formula gives the integrating factor |x-1|, which equals 1-x on -1 < x < 1. The alternative x-1 also works because it differs by a nonzero constant factor. Multiply both sides consistently to avoid a sign error.

What happens if I forget the integration constant?

In this problem, f(0)=0 fixes the integration constant at C=1. Omitting it effectively sets C=0 and gives f(0)=-1, violating the initial condition. It also changes the definite integral over [-1/2, 1/2] by -ln(3).

Is this a verified JEE 2025 previous-year question?

No, the supplied record labels the question 2025 but gives the date 2023-01-24. The year, date and shift remain unverified, and the supplied answer conflicts with the equation. Check the original paper before treating it as an official PYQ solution.