Let y=y(x) be the solution curve of the diffe | vedclass

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(B) The given differential equation is $\sin(2x^2) \ln(	an x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$.Dividing by $\sin(2x^2) \ln

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$1$

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(B) The given differential equation is $\sin(2x^2) \ln(	an x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$.Dividing by $\sin(2x^2) \ln(	an x^2)$,we get $dy + \frac{4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})}{\sin(2x^2) \ln(	an x^2)} dx = 0$.Rearranging,we recognize the derivative of a product: $d(y \ln(	an x^2)) = \frac{4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})}{\sin(2x^2)} dx$.Using $\sin(x^2 - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\sin x^2 - \cos x^2)$ and $\sin(2x^2) = 2 \sin x^2 \cos x^2$,the $RHS$ becomes $\frac{4x(\sin x^2 - \cos x^2)}{2 \sin x^2 \cos x^2} dx = 2x(\sec x^2 - \csc x^2) dx$.Integrating both sides: $y \ln(	an x^2) = \int 2x(\sec x^2 - \csc x^2) dx$.Let $t = x^2$,then $dt = 2x dx$. The integral becomes $\int (\sec t - \csc t) dt = \ln|\sec t + 	an t| - \ln|\csc t + \cot t| + C = \ln|	an(t/2)| + C$.Alternatively,using the identity $\int (\sec t - \csc t) dt = \ln|	an(t/2)| + C$,we find $y \ln(	an x^2) = \ln|	an(x^2/2)| + C$.Using the point $(\sqrt{\frac{\pi}{6}}, 1)$,we find $C$. At $x^2 = \frac{\pi}{6}$,$	an x^2 = \frac{1}{\sqrt{3}}$,so $\ln(1/\sqrt{3}) = \ln(	an(\pi/12)) + C$.At $x^2 = \frac{\pi}{3}$,$	an x^2 = \sqrt{3}$,so $y \ln(\sqrt{3}) = \ln(	an(\pi/6)) + C$.Solving these yields $|y| = 1$.

Let $y=y(x)$ be the solution curve of the differential equation $\sin(2x^2) \ln(\tan x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$ for $0 < x < \sqrt{\frac{\pi}{2}}$,which passes through the point $(\sqrt{\frac{\pi}{6}}, 1)$. Then $|y(\sqrt{\frac{\pi}{3}})|$ is equal to $.....$

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