<s>[INST] How to solve $y' = \sqrt {x+y+1}$ How does one solve $y' = \sqrt{x+y+1}$? I try substituting $v=x+y+1$ and using substitution methods, but it turned out to be so messy.
 [/INST] Another substitution is $x+y+1 = z^2$, so $1+y' = 2zz'$, from which $y' = 2zz'-1$:
$$
2zz'-1 = z \\
2zz' = z+1 \\
\frac {2zdz}{z+1} = dx \\
\int \left (1-\frac 1{z+1} \right ) dz = \frac 12\int dx \\
z - \ln \left |z+1 \right | = \frac x2 + C
$$
Starting this point, it makes sense to leave a solution as an implicit equation w.r.t $x,y$, so
$$
\sqrt{x+y+1} - \ln \left ( \sqrt{x+y+1}+1\right ) - \frac x2 = C
$$
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