Let $y=y(x)$ be the solution of the differential equation $$\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$$. If $y(1)=2$, then $y(2)$ is equal to :

I.F $=\mathrm{e}^{\int \frac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \frac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}$ <br/><br/>Put, $1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}$ <br/><br/>$\therefore$ $e^{\int-\frac{d t}{t}}=e^{-\ln t}=\frac{1}{t}=\frac{x^5}{1+x^5}$ <br/><br/>$$ \begin{aligned} y \cdot \frac{x^5}{1+x^5} & =\int \frac{x^5}{\left(1+x^5\right)} \times \frac{\left(1+x^5\right)^2}{x^7} d x \\\\ & =\int x^3 d x+\int x^{-2} d x \end{aligned} $$ <br/><br/>$y \cdot \frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c$ <br/><br/>$\text { Given that: } x=1 \Rightarrow y=2$ <br/><br/>$$ \begin{aligned} & 2 \cdot \frac{1}{2}=\frac{1}{4}-1+\mathrm{c} \\\\ & \mathrm{c}=\frac{7}{4} \\\\ & \mathrm{y} \cdot \frac{\mathrm{x}^5}{1+\mathrm{x}^5}=\frac{\mathrm{x}^4}{4}-\frac{1}{\mathrm{x}}+\frac{7}{4} \\\\ & \text { Now put, } \mathrm{x}=2 \\\\ & \mathrm{y} \cdot\left(\frac{32}{33}\right)=\frac{21}{4} \\\\ & \mathrm{y}=\frac{693}{128} \end{aligned} $$