If y = y(x) satisfies the differential equation 16 ( x + 9 x ) ( 4 + 9 + x ) cos y ? d y = ( 1 + 2 sin y ) ? d x , x 0 and y ( 256 ) = 2 , y ( 49 ) = , then 2 sin is equal to_____. [2026]

Question

If $y = y (x)$ satisfies the differential equation $16 (\sqrt{x + 9 \sqrt{x}}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 \sin α$ is equal to_____. [2026]

Smart Achievers · Accepted Answer

(4)

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

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

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

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

$\frac{1}{2} l n$ $| 1 + 2 \sin y |$ $= \frac{1}{4} l n | 4 + \sqrt{9 + \sqrt{x}} | + C$

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

$Substituting (256, \frac{π}{2})$

$\frac{1}{2} l n 3 = \frac{1}{2} l n 3 + C$ $C = 0$

$Substituting (49, α)$

$\frac{1}{2} l n (2 \sin α + 1) = \frac{1}{4} l n 8$

$l n (2 \sin α + 1) = \frac{1}{2} l n 8$

$l n (2 \sin α + 1) = l n (2 \sqrt{2})$

$2 \sin α + 1 = 2 \sqrt{2}$

$2 \sin α = 2 \sqrt{2} - 1$

Solution

Q.
If $y = y (x)$ satisfies the differential equation $16 (\sqrt{x + 9 \sqrt{x}}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 \sin α$ is equal to_____. [2026]

1 $3 (\sqrt{2} - 1)$

2 $\sqrt{2} - 1$

3 $2 (\sqrt{2} - 1)$

4 $2 \sqrt{2} - 1$

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Solution

Q. If y=y(x) satisfies the differential equation 16(x+9x)(4+9+x)cosy dy=(1+2siny) dx, x>0 and y(256)=π2, y(49)=α, then 2 sinα is equal to_____. [2026]

1 3(2-1)

2 2-1

3 2(2-1)