This ode is an example where \(y\) does not appear explicitly in the ode so not possible to directly solve for \(y\). It is given here to show possible problems with this method.\begin {equation} y^{\prime }=\sqrt {1+x+y} \tag {1A} \end {equation} This ode is squared to first solve for \(y\) which gives\begin {equation} \left ( y^{\prime }\right ) ^{2}=1+x+y \tag {2A} \end {equation} However, here care is needed. To get back to original ode (1A) then (2A) means two possible equations\[ y^{\prime }=\pm \sqrt {1+x+y}\] Hence the solutions obtained using (2A) can be the solution to one of these\begin {align} y^{\prime } & =+\sqrt {1+x+y}\tag {B1}\\ y^{\prime } & =-\sqrt {1+x+y} \tag {B2} \end {align}
Therefore the solution obtained by squaring both sides of (1A), which is done in order to solve for \(y\), must be checked to see if it satisfies the original ode, else it will be extraneous solution resulting from squaring both sides of the ode.
Starting from (2A), in normal form (by replacing \(y^{\prime }\) with \(p\)) it becomes\begin {align} y & =-x-1+p^{2}\tag {1}\\ & =xf+g\nonumber \end {align}
Where \(f=-1,g=-1+p^{2}\). Taking derivative w.r.t. \(x\) gives \begin {align} p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {2} \end {align}
Since \(\frac {\partial \phi }{\partial x}=-1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives the singular solution\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} But this solution does not satisfy the ode, hence it is extraneous. The general solution is found by finding \(p\) from (2). Since (2) is nonlinear, then it is inverted which gives\begin {align*} \frac {p+1}{2p} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {2p}{p+1} \end {align*}
Which is linear in \(x\). Solving gives\begin {equation} x=2p-2\ln \left ( p+1\right ) +c_{1} \tag {4} \end {equation} Instead of inverting this to find \(p\) in terms of \(x\), \(p\) is found from (1) which gives\begin {align*} y+x+1 & =p^{2}\\ p & =\pm \sqrt {y+x+1} \end {align*}
Substituting these solutions in (4) gives implicit solutions as\begin {align*} x & =2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\\ x & =-2\sqrt {y+x+1}-2\ln \left ( 1-\sqrt {y+x+1}\right ) +c_{1} \end {align*}
But only the first one above satisfies the ode. The second is extraneous. Therefore the final solution is\[ x=2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\] And no singular solutions exist. If instead of doing the above, \(p\) was found from (4) using inversion, then it will be\[ p=-\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1 \] Substituting this in (1) gives\[ y=-x-1+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1\right ) ^{2}\] But this general solution does not satisfy the original ode. In general, it is best to avoid squaring both side of the ode in order to solve for \(y\) as this can generate extraneous solutions. Only use this method if the original ode is already given in the form where \(y\) shows explicitly.