Verify $(y-x)y' = y-x+8$ has an explicit solution $y = x+ 4 \sqrt{x+2}$ Ive solved this problem and both sides of the function don't appear to be equal to me. However the solution in the back of the book says it is a solution.  Am I not simplifying enough? Thanks for the help!
(y-x)y' = y-x+8     ;   y = x+ 4 sqrt(x+2)
                       y'=( 2/sqrt(x+2)) +1 

Then I plug in the values of y and y' into the function on the left.
(x+4*sqrt(x+2) -x)*(2/sqrt(x+2)) +1  = x+4*sqrt(x+2) -x +8


Step 1:
$$y = x+ 4 \sqrt{x+2} \implies y' = \dfrac{2}{\sqrt{x+2}}+1$$
Step 2:
$$(y-x)y' = (x+ 4 \sqrt{x+2} - x) \left(\frac{2}{\sqrt{x+2}}+1\right) = 4 \left(\sqrt{x+2}+2\right)$$
Step 3:
$$ y-x+8  = x+ 4 \sqrt{x+2}  - x + 8 = 4 \left(\sqrt{x+2}+2\right)$$
