## Thomas' Calculus 13th Edition

As we are given that $y=x^2+\int_{1}^{x} \dfrac{1}{t} dt$ Now, $y'=2x^{(2-1)}+(\dfrac{1}{x})=2x+(\dfrac{1}{x})$ and $y''=2x^{(1-1)}+(-1)x^{(-1+1)}=2-(\dfrac{1}{x^2})$ Apply initial conditions. $y(1)=1$ This implies that $y'(1)=2x+(\dfrac{1}{x})=2(1)+\dfrac{1}{(1)}=3$ This tells us that the differential equation is satisfied.