Question

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

Answer 1

Answer: 1

$\sin \left(2x^2\right)\ln \left(\tan x^2\right)dy+\left(4xy-4\sqrt{2}x\sin \left(x^2-\frac{\pi }{4}\right)\right)dx=0$
$\ln \left(\tan x^2\right)dy+\frac{4xydx}{\sin \left(2x^2\right)}-\frac{4\sqrt{2}x\sin \left(x^2-\frac{\pi }{4}\right)}{\sin \left(2x^2\right)}dx=0$
$d\left(y\cdot \ln \left(\tan x^2\right)\right)-4\sqrt{2}x\frac{\left(\sin x^2-\cos x^2\right)}{\sqrt{2}-2\sin x^2\cos x^2}dx=0$
$d\left(y\ln \left(\tan x^2\right)\right)-\frac{4x\left(\sin x^2-\cos x^2\right)}{\left(\sin x^2+{\cos }^2\right)-1}dx=0$
$\Rightarrow \int d\left(y\ln \left(\tan x^2\right)\right)+2\int \frac{dt}{t^2-1}=\int 0$
$\Rightarrow y\ln \left(\tan x^2\right)+2\cdot \frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|=c$
$y\ln \left(\tan x^2\right)+\ln \left(\frac{\sin x^2+\cos x^2-1}{\sin x^2+\cos x^2+1}\right)=c$
Put $y=1$ and $x=\sqrt{\frac{\pi }{6}}$
$1\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=c$
Now
$x=\sqrt{\frac{\pi }{3}}\Rightarrow y(\ln \sqrt{3})+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+3}\right)$
$y(\ln \sqrt{3})=\ln \left(\frac{1}{\sqrt{3}}\right)$
$\Rightarrow y=-1$
$|y|=1$