Answer
$y'+(\dfrac{2x^3}{1+x^4})y=1$
Work Step by Step
We have: $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$
Apply product rule:
Thus, we have: $y'=(-\dfrac{1}{2})[\dfrac{4x^3}{(\sqrt{1+x^4})^3}] \int^{x}_1\sqrt{1+t^4}dt+\dfrac{1}{\sqrt{1+x^4}} (\sqrt{1+x^4})$
or, $y'=(\dfrac{-2x^3}{1+x^4})[(\dfrac{1}{\sqrt{1+x^4}})\int^{x}_1\sqrt{1+t^4}dt]+1$
or, $y'+(\dfrac{2x^3}{1+x^4})y=1$
This is the required differential equation.
