Question
If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x + 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}})cosydy = (1 + 2$ siny)dx, $x > 0$ and $y(256) = \frac{\pi}{2},y(49) = \alpha$, then 2 $sin\alpha$ is equal to :
Options
Solution
$\int\frac{cosy}{1 + 2siny}dy = \int\frac{dx}{16(\sqrt{9\sqrt{x} + x})(4 + \sqrt{9 + \sqrt{x}})}$
$$\begin{matrix} & 4 + \sqrt{9 + \sqrt{x}} = t \\ & \frac{1}{2\sqrt{9 + \sqrt{x}}} \times \frac{dx}{2\sqrt{x}} = 1dx \end{matrix}$$
$$\begin{matrix} & \frac{1}{2}ln|1 + 2siny| = \int_{}^{}\ \frac{4dt}{16t} + C \\ & \left. \ \frac{1}{2}ln|1 + 2siny| = \frac{1}{4}\ln \right|\ 4 + \sqrt{9 + \sqrt{x}} + C \\ & \frac{1}{2}ln(2siny + 1) = \frac{1}{4}ln|4 + \sqrt{9 + \sqrt{x}}| + C \\ & \text{~}\text{Substituting}\text{~}\left( 256,\frac{\pi}{2} \right) \\ & \frac{1}{2}ln3 = \frac{1}{2}ln3 + C\ C = 0 \end{matrix}$$
Substituting $(49,\alpha)$
$$\begin{matrix} & \frac{1}{2}ln(2sin\alpha + 1) = \frac{1}{4}ln8 \\ & ln(2sin\alpha + 1) = \frac{1}{2}ln8 \\ & ln(2sin\alpha + 1) = ln2\sqrt{2} \\ & 2sin\alpha + 1 = 2\sqrt{2} \\ & 2sin\alpha = 2\sqrt{2} - 1 \end{matrix}$$
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