<s>[INST] 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

 [/INST] We have $y'=\frac{y-x+8}{y-x}$ and a possible solution $y=x+4\sqrt{x+2}$ which gives $y'=\frac{2}{\sqrt{x+2}}+1$. Plugging this information into the DFQ we get: $$\frac{2}{\sqrt{x+2}}+1=\frac{x+4\sqrt{x+2}-x+8}{x+4\sqrt{x+2}-x}$$ which simplifies to: $$\frac{2}{\sqrt{x+2}}+\frac{\sqrt{x+2}}{\sqrt{x+2}}=\frac{4\sqrt{x+2}+8}{4\sqrt{x+2}}$$ When you divide right hand side term by factor 4, you get an identity. Conclusion?
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