Question

Let $y=y(x)$ be the solution curve of the differential equation $\sin \left(2 x^2\right) \log _e\left(\tan x^2\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^2-\frac{\pi}{4}\right)\right) d x=0$, $0

Answer 1

$ \sin \left(2 x^2\right) \ln \left(\tan x^2\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^2-\frac{\pi}{4}\right)\right) d x=0$
$ \ln \left(\tan x^2\right) d y+\frac{4 x y d x}{\sin \left(2 x^2\right)}-\frac{4 \sqrt{2} x \sin \left(x^2-\frac{\pi}{4}\right)}{\sin \left(2 x^2\right)} d x=0$
$d\left(y \cdot \ln \left(\tan x^2\right)\right)-4 \sqrt{2} x \frac{\left(\sin x^2-\cos x^2\right)}{\sqrt{2}-2 \sin x^2 \cos x^2} d x=0$
$ d\left(y \ln \left(\tan x^2\right)\right)-\frac{4 x\left(\sin x^2-\cos x^2\right)}{\left(\sin x^2+\cos ^2\right)-1} d x=0 $
$ \Rightarrow \int d\left(y \ln \left(\tan x^2\right)\right)+2 \int \frac{d t}{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$