Verification of a general solution to d'Alembert equation

I solved a nonlinear differential equation (d'Alembert one) by hand. Mathematica gives the same answer.

But I am not able to get Mathematica to verify the solution due to branch cuts.

Any one knows of a trick to verify the solution?

I tried different assumptions but can't find the right set. I know the solution is correct, well, at least I get same solution, but wanted to see if there is a trick to verify the solution back into the ODE, that is all.

 ClearAll[y,x];
ode = y'[x] == Sqrt[1 + x + y[x]]
sol = DSolve[ode, y, x]


Tried

  Simplify[ode /. sol]
Assuming[Element[x, Reals] && x > 0, FullSimplify[ode /. sol]]


and so on. Not able to get True.

Appendix

hand solution

Solve \begin{align*} {\frac {\rm d}{{\rm d}x}}y \left( x \right) &=\sqrt {1+x+y \left( x \right) } \end{align*}

This is d'Alembert ODE. It has the form $$y \left( x \right) =x g(y'(x)) + f(y'(x))$$. where $$g$$ and $$f$$ are functions of $$y'(x)$$.

Solving for $$y \left( x \right)$$ from the above and keeping only real solutions for $$y \left( x \right)$$ and letting $$p=y'(x)$$ gives \begin{align*} y \left( x \right) &= {p}^{2}-x-1 \tag{1} \\ \end{align*} ODE (1) is now solved.

In this ODE $$g(y'(x)) = -1$$ and $$f(y'(x))={p}^{2}-1$$. Taking derivatives of (1) w.r.t. $$x$$ and remembering that $$p$$ is a function of $$x$$ results in \begin{align*} p &= -1+ \left(2\,p\right) \frac{ \mathop{\mathrm{d}p}}{\mathop{\mathrm{d}x}}\\ p+1 &= \left(2\,p\right) \frac{ \mathop{\mathrm{d}p}}{\mathop{\mathrm{d}x}}\tag{2}\\ \end{align*} The singular solution is found by setting $$\frac{ \mathop{\mathrm{d}p}}{\mathop{\mathrm{d}x}}=0$$, which implies that $$p$$ is a constant. From the above, this results in \begin{align*} p+1&=0\\ \end{align*}
Solving the above for $$p$$ gives \begin{align*} p&=-1\\ \end{align*}

Substituting $$-1$$ values in (1) gives the singular solution $$y \left( x \right) =-x$$ But this solution does not satisfy the ODE, hence will not be used

The general solution is found when $$\frac{ \mathop{\mathrm{d}p}}{\mathop{\mathrm{d}x}}\neq 0$$. Rewriting (2) as \begin{align*} \frac{ \mathop{\mathrm{d}p}}{\mathop{\mathrm{d}x}} &={\frac {p+1}{2\,p}}\\ \end{align*} Inverting the above gives \begin{align*} \frac{ \mathop{\mathrm{d}x}}{\mathop{\mathrm{d}p}} &=2\,{\frac {p}{p+1}} \end{align*}
$$x \left( p \right)$$ is now the dependent variable and $$p$$ as the independent variable. Now this ODE is solved for $$x \left( p \right)$$. Since $${\frac {\rm d}{{\rm d}p}}x \left( p \right) = 2\,{\frac {p}{p+1}}$$ then $$x \left( p \right) = \int{2\,{\frac {p}{p+1}} \mathop{\mathrm{d}p}} = 2\,p-2\,\ln \left( p+1 \right) + C$$ Solving (using the computer) for $$p$$ from the above in terms of $$x$$ gives \begin{align*} p &= - \mathrm{LambertW} \left( -{{\rm e}^{-{\frac {x}{2}}-1+{\frac {C_{{1}}}{2}}}} \right) -1\\ \end{align*} Substituting the above solution for $$p$$ in Eq (1) gives the general solution.

$$y \left( x \right) = \left( \mathrm{LambertW} \left( -{{\rm e}^{-{\frac {x}{2}}-1+{\frac {C_{{1}}}{2}}}} \right) \right) ^{2}+2\, \mathrm {LambertW} \left( -{{\rm e}^{-x/2-1+1/2\,C_{{1}}}} \right) -x$$

ps. LambertW is ProductLog

• As an aside, you can substitute y[x] == v[x] - x which turns the ODE into a separable equation. – Chip Hurst Jun 24 '20 at 22:53

There are a few issues to be clarified: appropriate form of our differential equation and appropriate domain of the solution. Both issues are not independent and they arise due to playing with a general form of DSolve without specification of apropriate initial conditions. In order to demonstrate that the equation is satisfied we define:

ode = y'[x]^2 == 1 + x + y[x]
sol = DSolveValue[ode, y, x] // Quiet


We have chosen this way of defining ode in order to avoid checking whether y'[x] == Sqrt[1 + x + y[x]] or y'[x] == -Sqrt[1 + x + y[x]]. (it is a simple exercise to observe that this issue comes out when an integration constant is specified). The solution depends on its variable x and its parameter c (which could be determined with the initial conditions given).

f[x_, c_] := -x + 2 ProductLog[-E^(-1 - x/2 - c/2)] + ProductLog[-E^(-1 - x/2 - c/2)]^2


and now with appropriate domain of definiton (x > -c we assume that the solution is real like the original question seemed to assume, although this point in unnecessary) we have

Simplify[ ode /. {y[x] -> f[x, c], y'[x] -> Derivative[1, 0][f][x, c]}, x > -c]

True


When we deal with special functions like the Lambert W function (ProductLog) we should use FullSimplify rather than Simplify, nonetheless we chose the latter to show that it is straightforward checking whether this differential equation is satisfied. QED

Example c=1

With[{c = 1},
Plot[ ReIm @ f[x, c], {x, -3/2, 1/2}, PlotStyle -> Thickness[0.008],
AxesOrigin -> {0, 0}, Evaluated -> True]]


We can see that the solution is real for $$x\geq-c=-1$$ and its derivative is negative as mentioned above.

For quite a similar problem consider examining How to plot an implicit solution of a differential equation? where we had to change the variables in order to go ahead with reasonable analysis.