<s>[INST] Using the substitution $p=x+y$, find the general solution of $dy/dx=(3x+3y+4)/(x+y+1)$. 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?
 [/INST] Since $p(x)=x+y(x)$ therefore $y(x)=p(x)-x$. Thus 
$$
dy/dx=y'(x)=p'(x)-1.
$$
So the new equation is
$$
\frac{dp}{dx}=p'(x)=\frac{1}{p(x)+1}+4=\frac{4p(x)+5}{p(x)+1}=\frac{4p+5}{p+1}.
$$
This equation is separable. Using the usual method
$$
\frac{p+1}{4p+5}dp=1 dx,
$$
integrating
$$
\frac{p}{4}-\frac{1}{16}\log(4p+5)+C=x
$$
Substituting $p(x)=x+y(x)$ we obtain the solution that is the same as in your book.
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