If y=y(x) satisfies the differential equation | vedclass

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Question

(A) Given the differential equation: $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$. Separating the variables, we get

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$2 \sqrt{2}-1$

Vedclass · Answer

(A) Given the differential equation: $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$. Separating the variables, we get: $\int \frac{\cos y}{1+2 \sin y} \, dy = \int \frac{dx}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})}$. Let $u = 1+2 \sin y$, then $du = 2 \cos y \, dy$, so $\int \frac{\cos y}{1+2 \sin y} \, dy = \frac{1}{2} \ln |1+2 \sin y|$. For the $RHS$, let $t = 4+\sqrt{9+\sqrt{x}}$. Then $t-4 = \sqrt{9+\sqrt{x}}$. Squaring both sides: $(t-4)^2 = 9+\sqrt{x}$, so $\sqrt{x} = (t-4)^2 - 9$. Differentiating $t = 4+\sqrt{9+\sqrt{x}}$, we get $dt = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{dx}{4\sqrt{x(9+\sqrt{x})}} = \frac{dx}{4\sqrt{x+9\sqrt{x}}}$. Thus, $\frac{dx}{\sqrt{x+9\sqrt{x}}} = 4 \, dt$. The integral becomes: $\frac{1}{2} \ln |1+2 \sin y| = \int \frac{4 \, dt}{16t} = \frac{1}{4} \ln |t| + C = \frac{1}{4} \ln |4+\sqrt{9+\sqrt{x}}| + C$. Using $y(256) = \frac{\pi}{2}$: $\frac{1}{2} \ln(1+2 \sin \frac{\pi}{2}) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{256}}) + C \implies \frac{1}{2} \ln 3 = \frac{1}{4} \ln(4+\sqrt{9+16}) + C = \frac{1}{4} \ln 9 + C = \frac{1}{2} \ln 3 + C$. Thus $C = 0$. Now, for $y(49) = \alpha$: $\frac{1}{2} \ln(1+2 \sin \alpha) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{49}}) = \frac{1}{4} \ln(4+\sqrt{16}) = \frac{1}{4} \ln 8 = \frac{1}{4} \ln(2^3) = \frac{3}{4} \ln 2$. So $\ln(1+2 \sin \alpha) = \frac{3}{2} \ln 2 = \ln(2^{3/2}) = \ln(2\sqrt{2})$. Therefore, $1+2 \sin \alpha = 2\sqrt{2}$, which gives $2 \sin \alpha = 2\sqrt{2}-1$.

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

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