Answer
See below the result.
Work Step by Step
Given: $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$
Use product rule.
So, we get $y'=(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}$
and $x^2 y'=x^2[(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}]=-\int^{x}_1\dfrac{e^t}{t}dt+e^x$
or, $=-x[(\dfrac{1}{x})(\int^{x}_1\dfrac{e^t}{t}dt)]+(e^x)$
Thus, $x^2 y’=-xy+e^x$
Hence, this is our required differential equation.
