Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 0 down vote favorite
1

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.

share|improve this question
Did you try plugging it in and checking if it's a solution? It is. – Jim Feb 18 at 5:44
I mean how do we know it is another solution to the DE? – Idonknow Feb 18 at 5:47
how do you know your solution was general? – James S. Cook Feb 18 at 5:49
1  
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 at 5:52
Ooh, I think I do divide $y=-x$ in my solution. Thx for the guidance. – Idonknow Feb 18 at 6:20

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
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 at 6:00
up vote 1 down vote

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.

share|improve this answer
Nice contribution! + 1 Hello! ;-) – amWhy Feb 18 at 13:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.