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

$$\begin{gathered} \frac{dy}{dx}+\left(\frac{-3x^5{\tan }^{-1}{x}^3}{{(1+x^6)}^{\frac{3}{2}}}\right)y=2e^{\left\{\frac{x-\tan x}{\sqrt{1+x^6}}\right\}} \\ I.F.=e^{\int \frac{-3x^5{\tan }^{-1}{x}^3}{{(1+x^6)}^{\frac{3}{2}}}dx} \\ =e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}} \end{gathered}$$

Solution of differential equation

$$\begin{gathered} y.e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=\int 2x{e}^{\left(\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{1+x^6}}\right)}.e^{\left(\frac{{\tan }^{-1}\left(x^3\right)-x^3}{\sqrt{1+x^6}}\right)}dx \\ =\int 2xdx+c \\ y.e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=x^2+c \end{gathered}$$

Also it passes through origin
$$\begin{gathered} c = 0 \\ y\left(1\right).e^{\frac{\frac{\pi }{4}-1}{\sqrt{2}}}=1 \\ y\left(1\right).e^{\frac{\pi -4}{4\sqrt{2}}}=1 \\ y\left(1\right)=\frac{1}{e^{\frac{\pi -4}{4\sqrt{2}}}}=e^{\frac{4-\pi }{4\sqrt{2}}} \end{gathered}$$