## University Calculus: Early Transcendentals (3rd Edition)

$x^2y'=-xy+e^x$
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, $=-xy+e^x$