i have the differential equation $y'=\frac{y-x}{y-x+1}$, how i solve this?
try: i tryed to substitute $u=y-x$, then
$u=y-x\iff y=u+x\Rightarrow y'=u'+1$ then $y'=\frac{y-x}{y-x+1}$ become $u'+1=\frac{u}{u+1}$ and
$u'+1=\frac{u}{u+1}\\ \frac{du}{dx}=\frac{u}{u+1}-1\\ \frac{du}{dx}=1-\frac{1}{u+1}-1\\ \frac{du}{dx}=-\frac{1}{u+1}\\ (u+1)du=-dx\\ \displaystyle\int u+1du=-\int dx\\ \frac{1}{2}u^2+u=-x+C\\ u^2+2u=-2x+\overbrace{2C}^{A}\\ (y-x)^2+2(y-x)=-2x+A\\ y^2-2xy+x^2+2y-2x=-2x+A\\ y^2-2xy+x^2+2y=A\\ y^2-2xy+x^2+2y=A$
exists other way to solve this?
So, this ODE is an exact equation. We can write the DE as:
$$(y-x)dx+(x-y-1)dy=0$$
By exact equation, we mean that there is some function $F$ such that $dF=(y-x)dx+(x-y-1)dy$, i.e $dF=F_xdx+F_ydy=0$. We must also have the condition that the second mixed partial derivatives are equal...this will tell us that such a function $F$ actually exists. Thus,
$$\partial_y(y-x)=\partial_x(x-y-1) $$
This is equivalent to saying the the mixed partial derivatives of some well behaved function are equal. We can proceed by integrating:
$$F(x,y)=\int(y-x)dx=yx-\frac12x^2+\mu(y) $$
Then,
$$F_y=x+\mu'(y)=x-y-1 $$
$$\mu'(y)=-y-1$$ $$\mu(y)=-\frac12y^2-y+C $$
This means that
$$F(x,y)=yx-\frac12x^2-\frac12y^2-y+C=0 $$
Which is our desired result. You can solve for $y$ explicitly I believe.