$\sin \left(2 x^{2}\right) \log _{e}\left(\tan x^{2}\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)\right) d x=0$,$0 < x < \sqrt{\frac{\pi}{2}}$ નો ઉકેલ વક્ર $y=y(x)$ છે. જે બિંદુ $\left(\sqrt{\frac{\pi}{6}}, 1\right)$ માંથી પસાર થાય છે. તો $\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|=$ ..............
$\ln \left(\tan x^{2}\right) d y+\frac{4 x y d x}{\sin \left(2 x^{2}\right)}-\frac{4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)}{\sin \left(2 x^{2}\right)} d x=0$
$d\left(y \cdot \ln \left(\tan x^{2}\right)\right)-4 \sqrt{2} x \frac{\left(\sin x^{2}-\cos x^{2}\right)}{\sqrt{2}-2 \sin x^{2} \cos x^{2}} d x=0$
$d\left(y \ln \left(\tan x^{2}\right)\right)-\frac{4 x\left(\sin x^{2}-\cos x^{2}\right)}{\left(\sin x^{2}+\cos ^{2}\right)-1} d x=0$
$\int d\left(y \ln \left(\tan x^{2}\right)\right)+2 \int \frac{d t}{t^{2}-1}=\int 0$
$y \ln \left(\tan x^{2}\right)+2 \cdot \frac{1}{2} \ln \left|\frac{t-1}{t+1}\right|=c$
$y \ln \left(\tan x^{2}\right)+\ln \left(\frac{\sin x^{2}+\cos x^{2}-1}{\sin x^{2}+\cos x^{2}+1}\right)=c$
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 $x=\sqrt{\frac{\pi}{3}} \Rightarrow y(\ln \sqrt{3})+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+3}\right)$
$y(\ln \sqrt{3})=\ln \left(\frac{1}{\sqrt{3}}\right)$
$y =-1$
$|y|=1$
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વિધાન $1:$ $f(x)\, \le \,g\,(x)$ દરેક $x \in (0,\infty )$
વિધાન $2:$ $f(x)\, \le \,1$ દરેક $(x)\in (0,\infty )$ પરંતુ $g(x)\,\to \infty$ જો $x\,\to \infty$ હોય તો .