This looks like standard form of differential equations where we choose the substitution when we find the solution of system of equations of denominator and numerator, except there is square root on the right side and it complicates things here because:

since this system has no solutions we should use substitution $$z=x-y \Rightarrow z' = 1-y' \Rightarrow y'=1-z'$$

So the equation at the beginning becomes:

$$1-z'=\sqrt{\frac{z+1}{2-z}} \Rightarrow \frac{dz}{dx}=1-\sqrt{\frac{z+1}{2-z}} \Rightarrow dx=\frac{1}{1-\sqrt{\frac{z+1}{2-z}}}dz$$

And i am not really sure what to do from this point on. Integral that i need to compute is very complicated, even though i think it could be solved by implementing several substitutions but that is way to complicated so i am looking for some easier way to solve this differential equation.


You're almost there. Taking from where you left off, you can simplify the equation to

$$dx=\frac{\sqrt{2-z}}{\sqrt{2-z}-\sqrt{z+1}}dz$$ Then you multiply top and bottom by $\sqrt{2-z}+\sqrt{z+1}$, yielding $$dx=\frac{(2-z)+\sqrt{(2-z)(z+1)}}{3}dz$$ Can you take it from there?


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