Let y = y (x) be the solution curve of the differential equation (2...

Given differential equation, $\sin \left(2{x}^{2}\right)\ln \left(tan{x}^{2}\right)dy + \left(4xy - 4\sqrt{2}xsin\left({x}^{2} - \dfrac{\pi}{4}\right)\right)dx = 0$

$⇒ \ln \left(tan{x}^{2}\right)dy + \dfrac{4xydx}{\sin \left(2{x}^{2}\right)} - \dfrac{4\sqrt{2}xsin\left({x}^{2} - \dfrac{\pi}{4}\right)}{\sin \left(2{x}^{2}\right)}dx = 0$

$⇒ d\left(y \cdot \ln \left(tan{x}^{2}\right)\right) - 4\sqrt{2}x\dfrac{\left(sin{x}^{2} - cos{x}^{2}\right)}{\sqrt{2}\left(2sin{x}^{2}cos{x}^{2}\right)}dx = 0$

$⇒ d\left(yln\left(tan{x}^{2}\right)\right) - \dfrac{4x\left(sin{x}^{2} - cos{x}^{2}\right)}{{\left(sin{x}^{2} + \cos^{2}x\right)}^{2} - 1}dx = 0$

Now integrating both side,

$⇒ ∫ d\left(yln\left(tan{x}^{2}\right)\right) - ∫ \dfrac{4x\left(sin{x}^{2} - cos{x}^{2}\right)}{{\left(sin{x}^{2} + \cos^{2}x\right)}^{2} - 1}dx = ∫ 0$

Now let $sin{x}^{2} + \cos^{2}x = t ⇒ - 2x\left(sin{x}^{2} - \cos^{2}x\right) = dt$

$⇒ ∫ d\left(yln\left(tan{x}^{2}\right)\right) + 2 ∫ \dfrac{dt}{{t}^{2} - 1} = ∫ 0$

$⇒ yln\left(tan{x}^{2}\right) + 2 \cdot \dfrac{1}{2}\ln \left|\dfrac{t - 1}{t + 1}\right| = c$

$yln\left(tan{x}^{2}\right) + \ln \left(\dfrac{sin{x}^{2} + cos{x}^{2} - 1}{sin{x}^{2} + cos{x}^{2} + 1}\right) = c$

Put y = 1 and $x = \sqrt{\dfrac{\pi}{6}}$

$1\ln \left(\dfrac{1}{\sqrt{3}}\right) + ln\dfrac{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} - 1\right)}{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} + 1\right)} = c$

Now at $x = \sqrt{\dfrac{\pi}{3}}$

$⇒ y\left(ln\sqrt{3}\right) + ln\dfrac{\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2} - 1\right)}{\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2} + 1\right)} = \ln \left(\dfrac{1}{\sqrt{3}}\right) + \ln \left(\dfrac{\sqrt{3} - 1}{\sqrt{3} + 3}\right)$

$⇒ y\left(ln\sqrt{3}\right) = \ln \left(\dfrac{1}{\sqrt{3}}\right)</p><p>$⇒ y = - 1$</p><p>So, \left|y\right| = 1$