#### 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 $y(x_0)=y_0$

Published by
Pearson

ISBN 10:
0-32187-896-5

ISBN 13:
978-0-32187-896-0

$y'=x+y$;
$y(x_0)=y_0$

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