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I was trying to solve the differential equation :

$$2(x+y)dy +(3x+3y-1)dx = 0$$ using the substitution $s =x+y$ hence $s'= y'+1 $. This, however, didn't prove useful at all and I can still see no viable solution. Are there other ways to do it?

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You are almost done. First, note that the differential equation can be written as

$$\dfrac{dy}{dx}=-\dfrac{3(x+y)-1}{2(x+y)}.$$

Now use

$$z=x+y \implies \dfrac{dz}{dx}=1+\dfrac{dy}{dx}\implies \dfrac{dy}{dx}=\dfrac{dz}{dx}-1$$

for the differential equation

$$\dfrac{dz}{dx}-1 =-\dfrac{3z-1}{2z} \implies \dfrac{dz}{dx}=\dfrac{2z-3z+1}{2z}$$

$$\implies \dfrac{dz}{dx}=\dfrac{1-z}{2z} \implies \dfrac{2z}{1-z}dz=dx$$

Integrate the last expression and you are done. Can you complete it from here? This might be useful for the integration

$$\dfrac{2z-2+2}{1-z}dz=dx\implies \left[-2+\dfrac{2}{1-z} \right]dz=dx.$$

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YouMath 's ansswer is the best

You can try to solve it with exactness too

$$(3x+3y-1)dx +2(x+y)dy = 0$$

$$\int \frac {d\mu}{\mu} =\int \frac {dz}{1-z}$$

Where $z=x+y$ and $\mu (z) $ is the integrating factor

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