Differential Equations JEE 2026: Option D and the Mismatch

What is the supplied answer to this differential equations JEE 2026 question?

Option D is the supplied official answer, but the printed coefficient does not support the solution’s first simplification. This differential equations JEE 2026 worked answer reproduces the supplied method and identifies the mismatch. It does not validate the equation as printed.

The supplied record identifies this as JEE Main, January 2026, Slot 1. “Hard” is the question bank’s difficulty rating, and 180 seconds is its expected solving time, not an official time limit.

The function satisfies the supplied equation on the stated domain:

16 (\sqrt{x} + 9 \sqrt{x}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0 .

Use the given value to determine the requested quantity:

y (256) = \frac{π}{2}, y (49) = α; find 2 \sin α .

The options are:

The integration and endpoint calculations below follow the supplied reduced equation. The first step shows why that equation cannot be obtained from the printed coefficient.

Which substitution simplifies the nested roots, and does the cancellation work?

Use the entire outer radical as the new variable. This removes the nested roots, but the literal coefficient leaves an extra factor that the official reduction does not contain. Check the transformed coefficient before separating variables.

Set:

t = \sqrt{9 + \sqrt{x}}, \sqrt{x} = t^{2} - 9, x = (t^{2} - 9)^{2} .

Differentiate explicitly:

\frac{d x}{d t} = 2 (t^{2} - 9) \cdot 2 t = 4 t (t^{2} - 9), d x = 4 t (t^{2} - 9) d t .

The domain guarantees that the common factor is nonzero, so cancelling it is valid:

x > 0 ⟹ t > 3 ⟹ t^{2} - 9 > 0 .

Now check the printed bracket:

\sqrt{x} + 9 \sqrt{x} = 10 \sqrt{x} .

The literal equation therefore becomes:

160 (t^{2} - 9) (t + 4) \cos y d y = 4 t (t^{2} - 9) (1 + 2 \sin y) d t .

Cancelling gives:

40 (t + 4) \cos y d y = t (1 + 2 \sin y) d t .

By contrast, the supplied official reduction is:

4 (t + 4) \cos y d y = (1 + 2 \sin y) d t .

To obtain that reduction, the original coefficient multiplying the trigonometric differential would need to be:

16 \sqrt{x} \sqrt{9 + \sqrt{x}} (4 + \sqrt{9 + \sqrt{x}}) .

This is the coefficient consistent with the reduction, not a verified correction to the question. The remaining steps use the supplied official reduced equation, not the literal transcription.

How do you separate variables without losing either integration factor?

Keep the one-half from the left-hand substitution and the one-quarter from the right-hand coefficient. Losing either changes the power obtained after exponentiation. Starting from the official reduced equation, separate as follows:

\frac{\cos y}{1 + 2 \sin y} d y = \frac{d t}{4 (t + 4)} .

For the left side, use:

u = 1 + 2 \sin y, d u = 2 \cos y d y .

Thus:

\int \frac{\cos y}{1 + 2 \sin y} d y = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} \ln | u | .

Integrating both sides gives:

\frac{1}{2} \ln | 1 + 2 \sin y | = \frac{1}{4} \ln | t + 4 | + C .

Multiply by two and rename the arbitrary constant. The transformed domain makes the right-hand logarithm’s argument positive:

\ln | 1 + 2 \sin y | = \frac{1}{2} \ln (t + 4) + C_{1} .

Exponentiation yields:

| 1 + 2 \sin y | = A \sqrt{t + 4}, A > 0 .

At the supplied initial value:

1 + 2 \sin \frac{π}{2} = 3 > 0 .

Select the positive branch:

1 + 2 \sin y = K \sqrt{t + 4}, K > 0 .

Division during separation excludes the zero branch: $1 + 2 \sin y = 0 .$

That branch cannot satisfy the supplied initial value. It is therefore not a lost candidate for this problem.

How does the initial value lead to option D?

The initial value fixes the constant at one, and the target input gives option D. This follows from the supplied official reduced equation, pending verification of the original coefficient. The initial value fixes an endpoint of the implicit solution branch; it does not establish a finite derivative there.

At the initial input:

x = 256, \sqrt{x} = 16, t = \sqrt{9 + 16} = \sqrt{25} = 5, \sqrt{t + 4} = 3 .

Substitute the initial output into the integrated relation:

y = \frac{π}{2}, 1 + 2 \sin y = 3, 3 = 3 K ⟹ K = 1 .

Now evaluate the target endpoint:

x = 49, \sqrt{x} = 7, t = \sqrt{9 + 7} = \sqrt{16} = 4, \sqrt{t + 4} = \sqrt{8} = 2 \sqrt{2} .

Therefore:

1 + 2 \sin α = 2 \sqrt{2}, 2 \sin α = 2 \sqrt{2} - 1 .

That is option D. An endpoint-ratio check removes the constant directly:

\frac{1 + 2 \sin α}{3} = \sqrt{\frac{4 + 4}{5 + 4}} = \sqrt{\frac{8}{9}} .

Multiplying by three reproduces the same relation. This checks the endpoint arithmetic, not the disputed coefficient.

How can a bracket-division error produce option C?

$2 \sin α = 2 \sqrt{2} - 1 .$ This correctly obtained relation already answers the question. Option C can result from dividing the numerator incorrectly while trying to calculate the sine alone. The following step is wrong:

\sin α = \frac{2 \sqrt{2} - 1}{2} = \sqrt{2} - 1 (incorrect) .

Multiplying that mistaken result by two gives option C: $2 (\sqrt{2} - 1) .$

The correct division is:

\frac{2 \sqrt{2} - 1}{2} = \frac{2 \sqrt{2}}{2} - \frac{1}{2} = \sqrt{2} - \frac{1}{2} .

Both numerator terms must be divided by two. This is a failure to distribute division across the whole numerator, not a failure of separation of variables.

Stop once the requested doubled sine is isolated. Calculating the sine separately creates an unnecessary opportunity for error.

Which two practice questions check the same method?

Use these two teacher-created practice questions, not verified JEE past-paper questions, to check the integration factor and the additive one that must survive until the final subtraction. Attempt each before reading its worked answer. Neither depends on the disputed nested-root coefficient.

Question 1: Find the requested endpoint value:

\frac{d y}{d x} = \frac{1 + y}{2 (x + 1)}, x > - 1, y (0) = 1; find y (8) .

Separate and integrate:

\frac{d y}{1 + y} = \frac{d x}{2 (x + 1)}, \ln | 1 + y | = \frac{1}{2} \ln (x + 1) + C .

The initial value selects the positive branch and fixes its multiplier:

1 + y = K \sqrt{x + 1}, 2 = K \sqrt{1} ⟹ K = 2 .

Hence:

1 + y = 2 \sqrt{x + 1}, y (8) = 2 \sqrt{9} - 1 = 5 .

Question 2: Find the doubled sine at the specified input:

4 (x + 4) \cos y d y = (1 + 2 \sin y) d x, y (0) = 0; find 2 \sin (y (5)) .

Separate and use the same substitution:

\frac{\cos y}{1 + 2 \sin y} d y = \frac{d x}{4 (x + 4)}, u = 1 + 2 \sin y, d u = 2 \cos y d y .

Integration and the initial value give:

\frac{1}{2} \ln | 1 + 2 \sin y | = \frac{1}{4} \ln | x + 4 | + C, 1 + 2 \sin y = K \sqrt{x + 4} .

$1 = 2 K ⟹ K = \frac{1}{2} .$ At the target input:

1 + 2 \sin (y (5)) = \frac{1}{2} \sqrt{9} = \frac{3}{2}, 2 \sin (y (5)) = \frac{1}{2} .

On your next attempt, mark the one-half from integration and retain the additive one until the final subtraction. Check those two lines before checking the option.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to this differential equations JEE 2026 question?

Option D, 2√2 − 1, is the supplied official answer. It follows from the supplied reduced equation, but the printed coefficient does not support that reduction. The worked solution therefore reproduces the supplied method without validating the equation as printed.

Which substitution simplifies the nested roots in this question?

Use t = √(9 + √x), so √x = t² − 9 and dx = 4t(t² − 9) dt. Since x > 0, the factor t² − 9 is nonzero and can be cancelled. The printed equation then gives 40(t + 4) cos y dy = t(1 + 2 sin y) dt, which differs from the supplied reduction.

Where do the one-half and one-quarter factors come from during integration?

In the supplied reduced equation, substituting u = 1 + 2 sin y gives du = 2 cos y dy, producing the one-half on the left. Separating the right side gives dt/[4(t + 4)], producing the one-quarter. Together they lead to ln|1 + 2 sin y| = ½ ln(t + 4) + C.

Why can I get option C instead of option D?

Option C can result from incorrectly simplifying (2√2 − 1)/2 as √2 − 1. The correct division gives √2 − ½ because both numerator terms must be divided by two. Under the supplied reduction, stop at 2 sin α = 2√2 − 1, since that is already the requested quantity.