If y=y(x) satisfies the differential equation 16(x+9x)(4+√9+x) ydy=(1+2 y)dx,x0 and y(256…

Question

If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})\cos ydy=(1+2\sin y)dx,x>0$ and $y(256)=\frac{\pi }{2},y(49)=\alpha $, then $2\sin \alpha $ is equal to :

Accepted Answer

$\begin{gathered} \int \frac{\cos y}{1+2\sin y}dy=\int \frac{1}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})}dx \ \Rightarrow \frac{1}{2}\int \frac{2\cos y}{1+2\sin y}dy=\int \frac{1}{4t}dt \ 4+\sqrt{9+\sqrt{x}}=t \ \Rightarrow \frac{1}{2\sqrt{9+\sqrt{x}}}\cdot \frac{1}{2\sqrt{x}}dx=dt\Rightarrow \frac{1}{4\sqrt{x+9\sqrt{x}}}dx=dt \ \Rightarrow \frac{1}{2}\log |1+2\sin y|=\frac{1}{4}\log |t|+\log c\Rightarrow \sqrt{1+2\sin y}=c{t}^{1/4}\Rightarrow \frac{1}{4}\log |t|+\log c \ \Rightarrow \sqrt{1+2\sin y}=c{t}^{1/4}\Rightarrow \sqrt{1+2\sin y}=c(4+\sqrt{9+\sqrt{x}}{)}^{1/4} \ (256,\frac{\pi }{2})\Rightarrow \sqrt{3}=c(\sqrt{3})\Rightarrow c=1 \ (49,\alpha )\Rightarrow \sqrt{1+2\sin \alpha }=(1){8}^{1/4} \ \Rightarrow 1+2\sin \alpha =2\sqrt{2}\Rightarrow 2\sin \alpha =2\sqrt{2}-1 \end{gathered}$

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