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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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2 Answers 2

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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.

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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.

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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
    
@Vic, WA is agreeing with your incorrect answer because you're making your mistake, then feeding it to WA. If you give it the original equation: wolframalpha.com/input/… and ask it to solve it spits out a solution with the Lambert function which I think can by manipulated to look like the answer in your book. –  in_wolframAlpha_we_trust Jul 19 '12 at 10:35
    
@Vic: I already told you how to advance with the problem in my answer. It always happens when you solve a differential equation you get different forms of closed form solutions. But you can bring one form to the other. –  Mhenni Benghorbal Jul 19 '12 at 10:52

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