Let $y=y\left(x\right)$ be the solution of the differential equation. $\frac{dy}{dx}+\frac{5}{x(x^5+1)}y=\frac{{(x^5+1)}^2}{x^7},x>0$. If $y\left(1\right)=2,$then $128.y\left(2\right)$ is_____________

The given diff. equ. is of the form $\frac{dy}{dx}+Py=Q$

So, I.F$=e^{\int Pdx}=e^{\int \frac{5dx}{x(x^2+1)} }=e^{\int \frac{5x^{-6}dx}{(x^{-5}+1)}}$

Put,$1+x^{-5}=t\Rightarrow -5x^{-6}dx=dt$

$\Rightarrow e^{\int \frac{-dt}{t}}=e^{-\log t}=\frac{1}{t}=\frac{x^5}{1+x^5}$

Now, Solution is y (I.F)$=\int Q(I.F)+C$

$\Rightarrow y.\frac{x^5}{1+5^2}=\int \frac{x^5}{(1+x^5)}\times \frac{{(1+x^5)}^2}{x^7}dx$

$\Rightarrow \int x^{3}dx+\int x^{-2}dx$

$\Rightarrow y.\frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c$

$\Rightarrow 2.\frac{1}{2}=\frac{1}{4}-1+c [\because y\left(1\right)=2]$

$\Rightarrow c=\frac{7}{4}$

So,  $y.\frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+\frac{7}{4}$

Then  $y\left(2\right)\Rightarrow y.\left(\frac{32}{33}\right)=\frac{21}{4}\Rightarrow y=\frac{693}{128}$