Using the substitution $p=x+y$, find the general solution of $\frac{dy}{dx}=(3x+3y+4)/(x+y+1)$ Here are my steps:
Since $p=x+y$, $\frac{3x+3y+4}{x+y+1}=\frac{3p+4}{p+1}=\frac{1}{p+1}+3$
Therefore, integrate both sides $y=\ln(p+1)+3p+c$
$y=\ln(x+y+1)+3(x+y)+c$
But the answer in my book is $x+y-\frac{1}{4}\ln(4x+4y+5)=4x+c$ Is that correct?