## Thomas' Calculus 13th Edition

$y'=x+y$; $y(x_0)=y_0$
Here, $y'=-1+(1+x_0+y_0)e^{(x-x_0)}$ and $x+y=x+[-1-x+(1+x_0+y_0)e^{(x-x_0)}]$ This implies that $x+y=-1+(1+x_0+y_0)e^{(x-x_0)}$ Thus, this proves that $y'=x+y$ and $y(x_0)=y_0$