Answer
$x^2y'=-xy+e^x$
Work Step by Step
since, we have $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$
Apply product rule.
Thus, we have $y'=(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}$
Then $x^2 y'=x^2[(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}]=-\int^{x}_1\dfrac{e^t}{t}dt+e^x$
or, $=-x[(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt]+e^x$
or, $=-xy+e^x$