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I though about this trick and then found an example to apply it to: $$\frac{dy}{dx}=1+\frac{2}{x+y}$$ This is the trick: add $\frac{dx}{dx}=1$ to both parts $$\frac{dy}{dx}+\frac{dx}{dx}=1+\frac{2}{x+y}+1$$ Using the linearity of $d$ $$\frac{d(x+y)}{dx}=2\frac{1+(x+y)}{x+y}$$ $$\frac{(x+y)d(x+y)}{1+(x+y)}=2dx$$ $$d(x+y)-\frac{d(x+y)}{1+(x+y)}=2dx$$ $$-\frac{d(x+y)}{1+(x+y)}=2dx-d(x+y)$$ Now $2dx-d(x+y)=2dx-dx-dy=dx-dy=d(x-y)$ $$-\frac{d(x+y)}{1+(x+y)}=d(x-y)$$ $$\frac{d(1+x+y)}{1+(x+y)}=d(y-x)$$ Integrating: $$\ln|1+x+y|=(y-x)+\ln C$$ $$1+x+y=C\exp\left(y-x\right)$$ Is this a one-off case, or a particular example of a certain method? Does anyone know more examples of ODE's that can be solved similarly? I know the integrating multiplier theory quite well, but this one seems like something extra to that.

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I can't tell what happened when you went from dx to d(y-x). – Phira Jun 2 '12 at 12:55
I'll insert the missing steps – Valentin Jun 2 '12 at 13:01
Thanks for explaining. – Phira Jun 2 '12 at 13:18
You are making a change of variables. Your new variable is $x+y$ rather than the original $y$. This substitution is suggested by the form of the right-hand side. – GEdgar Jun 2 '12 at 13:34
It's a nice idea, and aas you can see will work for quite a few variants. Your calculation was more messy than necessary. Let $u=x+y$. Then your first line can be rewritten as $\frac{du}{dx}= 2+\frac{2}{u}$. This is a separable DE, solve for $u$. This would have made the later calculations shorter and more transparent. – André Nicolas Jun 2 '12 at 14:00
up vote 0 down vote accepted

In general for equations of the form $y'(x)=f(ax+by+c)$, (where $a, b$ and $c$ are constantes $b\neq 0$), you can use the change of variables $u=ax+by+c$. You obtain the equation: $$ y'=\frac{u'-a}{b}=f(u) $$ which can be solved using separation of variables: $$ \frac{du}{b \,f(u)+a}=dx $$

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