solve $y'=\frac{2x+y-1}{x+y+2}$ I need some help solving this differential equation, I'm doing everything as I was taught but I just need help at the last step, I can't seem to get this right. the ode we want to solve is $$y'=\frac{2x+y-1}{x+y+2}$$
if $2x+y-1=0$ and $x+y+2=0$ then $x=3$ and $y=-5$, we can now define $\tilde x = x-3$ and $\tilde y = y+5$  notice that $\tilde y' = y'$, and now we have a new differential equation:
$$\tilde y'= \frac{2\tilde x +6 +\tilde y-5-1}{\tilde x+3+\tilde y -5+2}=\frac{2\tilde x+\tilde y}{\tilde x+\tilde y}=\frac{2+\frac{\tilde y}{\tilde x}}{1+\frac{\tilde y}{\tilde x}}$$.
this is an homogenous ode that we can solve as one, define $z=\frac{ \tilde y}{\tilde x}$, and we have $z+\tilde xz'=\frac{2+z}{1+z}$
which simplifies to $\tilde x z' =\frac{2-z^2}{1+z}$. This is an ODE with separable variables, this simplifies to $\int \frac{1+z}{2-z^2}dz=\int \frac{1}{\tilde x}d \tilde x$
the RHS is simply $\ln |\tilde x|+c$
The LHS can be done using partial fractions, and the result i got was (after some simplification): $$-\frac{1}{2\sqrt{2}}\ln{(\frac{\sqrt 2-z}{\sqrt 2 +z})}-\frac{1}{2}\ln{(2-z^2)}=\ln |\tilde x|+c$$
And now I'm not sure what to do. It is possible that i misscalculated the LHS integral, but if I didn't and my result is correct, I'm not sure how to proceed. The best we can do in this case is give an implicit formula rather than a direct. The best I can come up with is transfer back from $z$ to $y$ and $x$ and give an answer implicitly.
Is that really the answer?

The solution on implicite form is shown below :

