# 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?

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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.
For the left hand side of your differential equation, you should substitute $\frac{dy}{dx} = \frac{dp}{dx} - 1\,,$ since $p = x + y$, and then advance with your solution by separation of variable method.
Thanks for answering. I'm trying to use the method you provided. But I'm wondering why the answer given by integrate both sides is wrong. WolframAlpha also choose to integrate both sides and give out $y=ln(x+y+1)+3(x+y)+c$. –  Vic. Jul 19 '12 at 10:29