Solving the exact differential equation y'=(x-y)/(x+y)

Question

I need to solve the following exact equation:

$y' = \frac{(x-y)}{(x+y)}$

I've been taught to put those in the form $M(x, y)dx + N(x, y)dy = 0$ and to make sure ${dM}\over{dy}$ = ${dN}\over{dx}$

So I transform the equation into $\frac{1}{(x+y)} dx + \frac{1}{(y-x)}dy=0$

However, $\frac{dM}{dy}=\frac{-1}{(y+x)^2}$ and $\frac{dN}{dx}=\frac{1}{(x-y)^2}$

They are not equal? I'm not supposed to use an integrating factor either.. so I'm a bit at a loss :(

The answer is supposed to be $x^2 -2xy -y^2 = c$

How do I get there?

Thanks a lot in advance :)

Answer 1

Try doing the algebra differently: $$-(x-y)dx+(x+y)dy=0\ .$$ Can you take it from here?

BTW if your instructor really wants you not to use an integrating factor, this is no good: we have multiplied both sides by $x+y$, which is an integrating factor even though we did not need to find it by any complicated method.

Comment. Since you have been told the solution, you could work out the method by "reverse-engineering" the answer: just take your solution $$x^2-2xy-y^2=C$$ and differentiate with respect to $x$ to get $$(2x-2y)dx-(2x+2y)dy=0\ ;$$ this gives you the hint as to how to start solving the DE.

Answer 2

I do not know if you are obliged to go through these steps; so forgive me if what I write is not an answer to your post.

The differential equation simplifies a lot if you define $x+y=z$; so $y=z-x$, $y'=z'-1$ and the rhs write $\frac {2x-z}{z}=\frac {2x}{z}-1$. Simplifying then leads to $$z z'=2x$$which, by integration of both sides, leads to $$z^2=2x^2+C$$ Back to the definition of $z$,$$(x+y)^2=2x^2+C$$ which, after development, leads to $$x^2 -2xy -y^2 = C$$