Given $(x^2 +3xy +y^2)dx - x^2 dy =0$ solve the DE.

Given $(x^2 +3xy +y^2)dx - x^2 dy =0$ solve the DE. I got the answer $\frac{x}{x+y}+ \ln|x|=C$ . But the answer provided got one extra answer , that is $y=-x$ can anyone explain to me why we need to include the extra answer.

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Did you try plugging it in and checking if it's a solution? It is. – Jim Feb 18 '13 at 5:44
I mean how do we know it is another solution to the DE? – Idonknow Feb 18 '13 at 5:47
how do you know your solution was general? – James S. Cook Feb 18 '13 at 5:49
How do you know it is another solution? Or how do you know there is another solution? You know it is another solution because you can simply test it and observe that it is a solution. How you can know if there is another solution or not is in general a much harder question. Since you solved the DE (as opposed to guessing a solution) then by looking back at your steps you should be able to find an assumption that you made that rules out $y = -x$. For example, as James suggests in his answer, did you at any point divide by or cancel a $(x + y)$ term from the equation? – Jim Feb 18 '13 at 5:52
Ooh, I think I do divide $y=-x$ in my solution. Thx for the guidance. – Idonknow Feb 18 '13 at 6:20

Explicit substitution of $y=-x$ into the given DEqn shows it is a soln. Notice: $$(x^2+3xy+y^2)dx-x^2dy = -x^2dx-x^2(-dx)=0.$$ Thus it is a solution. Why did your method miss it? Probably you divided by a term which is zero when $y=x$ so you lost that possibility right at that step.
I'm not sure what method you used, but to be clear, your answer has division by $x+y$ hence clearly it does not contain $y=-x$ except perhaps as some sort of limit. Usually the extraneous solution comes from division in separation of variables or from multiplication by an integrating factor which is zero along some locus. In contrast, the most famous integrating factor does not present this peculiarity as it is formed as an always nonzero exponential... so to answer your question more precisely you'll have to be more precise about how you "got it" – James S. Cook Feb 18 '13 at 6:00
Besides to @James's answer, we can see the problem as follows also: $$x^2\left(\left(1+\frac{y}x+\frac{y^2}{x^2}\right)dx-dy\right)=0$$ so if $x\neq0$ then by setting $u=y/x$ we get: $$(1+3u+u^2)dx-(udx+xdu)=0$$ This latter one is equal to $$\frac{du}{(u+1)^2}=\frac{dx}x,~~x\neq0$$ You certaily note that here that we should put $u\neq-1$ or $y\neq-x$ to find the solution above. But $y=-x$ itself is a solution as well (James did it). Now since this solution can not be given by that general solution so it isa singular solution.