Let the solution curve y=y(x) of the differential equation $\frac{\text{dy}}{\text{dx}}-\frac{3{\mathrm{x}}^5{\tan }^{-1}\left({\mathrm{x}}^3\right)}{{(1+{\mathrm{x}}^6)}^{3/2}}\mathrm{y}=2x \exp \left\{\frac{{\displaystyle {\mathrm{x}}^3-{\tan }^{-1}{\mathrm{x}}^3}}{{\displaystyle \sqrt{(1+{\mathrm{x}}^6)}}}\right\}$  pass through the origin. Then $y\left(1\right)$ is equal to:

$\begin{array}{l}{\text{IF = e}}^{\int \frac{-3{\mathrm{x}}^5{\tan }^{-1}\left({\mathrm{x}}^3\right)}{{(1+{\mathrm{X}}^6)}^{3/2}}\mathrm{dx}}\\ ={\mathrm{e}}^{\int \frac{-\mathrm{t}.\tan \text{ t}}{\sqrt{1+{\tan }^2\mathrm{t}}}\mathrm{dt}}\\ ={\mathrm{e}}^{\int \text{-t sin t dt}}\end{array}$

$\begin{array}{l}={\mathrm{e}}^{\text{t cos t-sin t}}\\ ={\mathrm{e}}^{\frac{\text{t-tan t}}{\sec \text{ t}}}={\mathrm{e}}^{\frac{{\text{tan}}^{-1}{\mathrm{x}}^3-{\mathrm{x}}^3}{\sqrt{1+{\mathrm{x}}^6}}}\end{array}$

${\text{y.e}}^{\frac{-({\mathrm{x}}^3-{\tan }^{-1}{\mathrm{x}}^3)}{\sqrt{1+{\mathrm{x}}^6}}}=\int 2\mathrm{x}.{\mathrm{e}}^{\frac{({\mathrm{x}}^3-{\tan }^{-1}{\mathrm{x}}^3)}{\sqrt{1+{\mathrm{x}}^6}}}{\mathrm{e}}^{\frac{-({\mathrm{x}}^3-{\tan }^{-1}{\mathrm{x}}^3)}{\sqrt{1+{\mathrm{x}}^6}}}\mathrm{dx}$

${\text{y.e}}^{\frac{{\displaystyle -({\mathrm{x}}^3-{\tan }^{-1}{\mathrm{x}}^3)}}{{\displaystyle \sqrt{1+{\mathrm{x}}^6}}}}=\int 2\mathrm{xdx}+\mathrm{c}$

$y\cdot e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=x^2+c$

$\mathrm{x}=0,\mathrm{y}=0\Rightarrow \mathrm{c}=0$

$y\left(1\right)\cdot e^{\frac{{\tan }^{-1}\left(1\right)-1}{\sqrt{2}}}=1$

$y\left(1\right)\cdot e^{\frac{\frac{\pi }{4}-1}{\sqrt{2}}}=1$
$\Rightarrow \mathrm{y}\left(1\right)={\mathrm{e}}^{\frac{4-\mathrm{\pi }}{4\sqrt{2}}}$