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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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    yes, I guess in the end it al boils down to changing the variable, though this "method" is meant suggest the change, or make it obvious. Thanks for your comments2012-06-02

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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 $