If y=y(x) satisfies the differential equation 16( x+9 x )(4+ 9+ x ) y ~d y=(1+2 y) d x, x>0 and y(256)=(π)/(2), y(49)=α , then 2 α is equal to :

Correct answer: C. 2 2 -1 ∫ ( y)/(1+2 y) d y=∫ d x 16( 9 x +x )(4+ 9+ x ) 4+ 9+ x =t 1 2 9+ x × d x 2 x =1 d x (1)/(2) n|1+2 y|=∫ (4 d t)/(16 t)+C .(1)/(2) n|1+2 y|=(1)/(4) n \, 4+ 9+ x +C (1)/(2) n(2 y+1)=(1)/(4) n|4+ 9+ x |+C Substituting (256, (π)/(2)) (1)/(2) n 3=(1)/(2) n 3+C C=0 Substituting (49, α) (1)/(2) n(2 α+1)=(1)/(4) n 8 n(2 α+1)=(1)/(2) n 8 n(2 α+1)= n 2 2 2 α+1=2 2 2 α=2 2 -1

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Differential Equations

If y=y(x) satisfies the differential equation 16( x+9 x )(4+ 9+ x )...

y=y(x)

satisfies the differential equation

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

and

y(256)=\frac{\pi}{2}, y(49)=\alpha

, then

2 \sin \alpha

is equal to :

Options

A $2(\sqrt{2}-1)$
B $3(\sqrt{2}-1)$
C $2 \sqrt{2}-1$
D $\sqrt{2}-1$

Correct answer

C. $2 \sqrt{2}-1$

Step-by-step solution

\int \frac{\cos y}{1+2 \sin y} d y=\int \frac{d x}{16(\sqrt{9 \sqrt{x}+x})(4+\sqrt{9+\sqrt{x}})}

4+\sqrt{9+\sqrt{x}}=t

\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{d x}{2 \sqrt{x}}=1 d x

\frac{1}{2} \ell n|1+2 \sin y|=\int \frac{4 d t}{16 t}+C

\left.\frac{1}{2} \ell n|1+2 \sin y|=\frac{1}{4} \ell n \right\rvert\, 4+\sqrt{9+\sqrt{x}}+C

<br>

\frac{1}{2} \ell n(2 \sin y+1)=\frac{1}{4} \ell n|4+\sqrt{9+\sqrt{x}}|+C

<br><br>Substituting \left(256, \frac{\pi}{2}\right)

\frac{1}{2} \ell n 3=\frac{1}{2} \ell n 3+C \quad C=0

Substituting

(49, \alpha)

\frac{1}{2} \ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{4} \ell \mathrm{n} 8

\ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{2} \ell \mathrm{n} 8

\ell \mathrm{n}(2 \sin \alpha+1)=\ell \mathrm{n} 2 \sqrt{2}

2 \sin \alpha+1=2 \sqrt{2}

2 \sin \alpha=2 \sqrt{2}-1

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