Let the solution curve y = y ( x ) of the differential equation...

Given differential equation is$\dfrac{dy}{dx} + \left(\dfrac{- 3{x}^{5}\tan^{- 1}\left({x}^{3}\right)}{{\left(1 + {x}^{6}\right)}^{3 / 2}}\right)y = 2x{e}^{\left\{\dfrac{{x}^{3} - \tan^{- 1}{x}^{3}}{\sqrt{{\left(1 + x\right)}^{6}}}\right\}}$

This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$.

Hence,

$I . F . = {e}^{∫ \dfrac{- 3{x}^{5}\tan^{- 1}{x}^{3}}{{\left(1 + {x}^{6}\right)}^{3 / 2}}dx}$

Put $\tan^{- 1}{x}^{3} = u ⇒ \left(\dfrac{3{x}^{2}}{1 + {x}^{6}}\right)dx = du$

$I . F . = {e}^{- ∫ \dfrac{utanu}{\sqrt{1 + \tan^{2}u}}du}$

$⇒ I . F . = {e}^{- ∫ \dfrac{utanu}{\sec u}du}$

$⇒ I . F . = {e}^{- ∫ usinudu}$

$⇒ I . F . = {e}^{- \left[- ucosu + ∫ cosudu\right]}$

$⇒ I . F . = {e}^{ucosu - \sin u}$

$⇒ I . F . = {e}^{\left\{\tan^{- 1}\left({x}^{3}\right)\left(\dfrac{1}{\sqrt{1 + {x}^{6}}}\right) - \left(\dfrac{{x}^{3}}{\sqrt{1 + {x}^{6}}}\right)\right\}}$

$⇒ I . F . = {e}^{\left\{\dfrac{\tan^{- 1}{x}^{3} - {x}^{3}}{\sqrt{1 + {x}^{6}}}\right\}}$

Solution of differential equation is

$y \cdot e\dfrac{\tan^{- 1}{x}^{3} - {x}^{3}}{\sqrt{1 + {x}^{6}}} = ∫ 2x{e}^{\left(\dfrac{{x}^{3} - \tan^{- 1}{x}^{3}}{\sqrt{1 + {x}^{6}}}\right)} \cdot {e}^{\left(\dfrac{\tan^{- 1}\left({x}^{3}\right) - {x}^{3}}{\sqrt{1 + {x}^{6}}}\right)}dx$

$⇒ y \cdot {e}^{\left\{\dfrac{\tan^{- 1}{x}^{3} - {x}^{3}}{\sqrt{1 + {x}^{6}}}\right\}} = ∫ 2xdx$

$⇒ y \cdot {e}^{\left\{\dfrac{\tan^{- 1}{x}^{3} - {x}^{3}}{\sqrt{1 + {x}^{6}}}\right\}} = {x}^{2} + C$

Also, it passes through the origin then $C = 0$, hence

$y{e}^{\left\{\dfrac{\tan^{- 1}{x}^{3} - {x}^{3}}{\sqrt{1 + {x}^{6}}}\right\}} = {x}^{2}$

Now, put $x = 1$, then we get

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

$⇒ y\left(1\right)e\dfrac{\pi - 4}{4\sqrt{2}} = 1$

$⇒ y\left(1\right) = \dfrac{1}{{e}^{\left(\dfrac{\pi - 4}{4\sqrt{2}}\right)}}$

$⇒ y\left(1\right) = e\dfrac{4 - \pi}{4\sqrt{2}}$