Let the solution curve $$y = y(x)$$ of the differential equation $$\frac{dy}{dx} - \frac{3x^5\tan^{-1}x^3}{1+x^{6\cdot 7}} y = 2x \exp\frac{x^3 - \tan^{-1}x^3}{(1+x)^7}$$ pass through the origin. Then $$y(1)$$ is equal to:

$$\text{IF} = e^{\int \frac{-3x^5 \tan^{-1}\left(x^3\right)}{\left(1 + x^6\right)^{3/2}} \, dx}$$

Using the substitution $$x^3 = \tan t$$, we have $$3x^2 \, dx = \sec^2 t \, dt$$:

$$= e^{\int \frac{-t \cdot \tan t}{\sqrt{1 + \tan^2 t}} \, dt}$$

$$= e^{\int -t \sin t \, dt}$$

$$= e^{t \cos t - \sin t}$$

$$= e^{\frac{t - \tan t}{\sec t}} = e^{\frac{\tan^{-1}x^3 - x^3}{\sqrt{1 + x^6}}}$$

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

$$y \cdot e^{\frac{-\left(x^3 - \tan^{-1}x^3\right)}{\sqrt{1 + x^6}}} = \int 2x \, dx + c$$

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

$$x = 0, y = 0 \implies c = 0$$

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

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

$$\implies y(1) = e^{\frac{4 - \pi}{4\sqrt{2}}}$$