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I am reading a book about differential equations and I found this equation in the chapter about homogeneous differential equations. The books solves this equation by turning it to a homogeneous.

This reminds me the kind of equation that can be solved this way.

$M(x,y) = x + y -1$ $N(x,y) = x-y+1$

$M_{y} = 1$ $N_{x} = 1$

so $\phi (x,y) = \int_{0}^{x}t+y-1 dt +\int_{0}^{y} -t -1 dt$

Is this thought correct?

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You could try it and see if you get the right answer. In principle, what you have found is that the equation is "exact", and the method for solving exact equations applies.

EDIT: in response to the comment, what I had in mind by "the method for solving exact equations" was, integrate $M$ with respect to $x$, getting $(1/2)x^2+xy-x+\phi(y)$ for some function $\phi$ to be determined, then differentiate that with respect to $y$, getting x+\phi'(y); compare this to $N$ to see \phi'(y)=-y+1, so $\phi(y)=-(1/2)y^2+y+C$ for some constant $C$, so we have $(1/2)x^2+xy-x-(1/2)y^2+y+C=0$ Is that what you get by your method?

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