Answer
$y'=x+y$;
$y(x_0)=y_0$
Work Step by Step
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 $y(x_0)=y_0$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 9: First-Order Differential Equations - Section 9.1 - Solutions, Slope Fields, and Euler's Method - Exercises 9.1 - Page 531: 21
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