Let y = y(x) be the solution of the differential equation (dy)/(dx)...

Given:

$\dfrac{dy}{dx} + \dfrac{5}{x\left(1 + {x}^{5}\right)}y = \dfrac{{\left(1 + {x}^{5}\right)}^{2}}{{x}^{7}}$

This is linear differential equation.

$I . F . = {e}^{∫ \dfrac{5}{x\left(1 + {x}^{5}\right)}dx}$

$⇒ I . F . = {e}^{∫ \dfrac{5{x}^{5}}{{x}^{6}\left(1 + {x}^{5}\right)}dx}$

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

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

$⇒ I . F . = {e}^{- ∫ \dfrac{d\left({x}^{- 5} + 1\right)}{\left({x}^{- 5} + 1\right)}}$

$⇒ I . F . = {e}^{- \log_{e}\left({x}^{- 5} + 1\right)}$

$⇒ I . F . = {\left({x}^{- 5} + 1\right)}^{- 1} = \dfrac{{x}^{5}}{\left({x}^{5} + 1\right)}$

So, solution is given by

$y\dfrac{{x}^{5}}{\left({x}^{5} + 1\right)} = ∫ \dfrac{{x}^{5}}{\left({x}^{5} + 1\right)} \times \dfrac{{\left({x}^{5} + 1\right)}^{2}}{{x}^{7}}dx$

$⇒ \dfrac{y{x}^{5}}{\left({x}^{5} + 1\right)} = ∫ \left(\dfrac{{x}^{5} + 1}{{x}^{2}}\right)dx$

$⇒ \dfrac{y{x}^{5}}{\left({x}^{5} + 1\right)} = ∫ \left({x}^{3} + \dfrac{1}{{x}^{2}}\right)dx$

$⇒ \dfrac{y{x}^{5}}{\left({x}^{5} + 1\right)} = \dfrac{{x}^{4}}{4} - \dfrac{1}{x} + C$

Since, $y\left(1\right) = 2$, so

$\dfrac{2}{2} = \dfrac{1}{4} - 1 + C ⇒ C = \dfrac{7}{4}$

So,

$\dfrac{y{x}^{5}}{\left({x}^{5} + 1\right)} = \dfrac{{x}^{4}}{4} - \dfrac{1}{x} + \dfrac{7}{4}$

Put $x = 2$, then we get

$\dfrac{y\left(2\right) \times 32}{33} = \dfrac{16}{4} - \dfrac{1}{2} + \dfrac{7}{4}$

$⇒ y\left(2\right) = \dfrac{21 \times 33}{128} = \dfrac{693}{128}$