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Given the ode

\begin{equation} y^{\prime}=\sqrt{1+x+y}\tag{1} \end{equation}

The question is, is this a D'Alembert's ode or not? As is known, D'Alembert's has the form Wikipedia

\begin{equation} y=xf\left( p\right) +g\left( p\right) \tag{2} \end{equation}

Where $p\equiv\frac{dy}{dx}=y^{\prime}$. As it stands (1) is not of the form (2). But after squaring both sides of (1) the ode becomes

\begin{align} \left( y^{\prime}\right) ^{2} & =1+x+y\nonumber\\ y & =-x-1+\left( y^{\prime}\right) ^{2}\tag{3}\\ & =xf\left( p\right) +g\left( p\right) \nonumber \end{align}

Where \begin{align*} f\left( p\right) & =-1\\ g\left( p\right) & =-1+p^{2}% \end{align*}

Now ode (3) is D'Alembert's. But this is after squaring both sides of (1).

So the question is, is it mathematically correct to say (1) is D'Alembert's ode or not?

Maple agrees and says that (1) is D'Alembert's ode. So it must have squared both sides.

restart;
ode:=diff(y(x),x)=sqrt(1+x+y(x));
DEtools:-odeadvisor(ode);

gives

 [[_homogeneous, `class C`], _dAlembert]

But some might not agree with Maple here.

I am not sure if one is allowed to do that, or if one must only apply the form comparison on the original ode without squaring it. Hence my question.

I am well aware that squaring the ode and solving the squared ode will introduce extraneous solutions to the original ode, and one must therefore verify that each solution generated satisfies the original ode, else removed as extraneous. But that is not my question.

My question is on the type of the original ode itself. In particular, I am not sure now if squaring both sides of the ode will generate an ode with all of its solutions that fail to verify the original ode.

I do not have such an example myself of such case, but this does not mean such case does not exist.

In the example given in this question at the top, squaring the ode generates two solutions, one of which is extraneous and can be discarded, but the second one does verify the original ode.

On a side note, without squaring the above ode and solving it as d'Alembert type, I now have no idea how to solve the original ode as is, as I do not know what type it is, and hence do not know what method to apply to it. Maple says the above ode is also of type _homogeneous, class C. But I looked at Maple's website, and read the description, but did not follow it and still have no idea still what it means as there is no actual reference outside Maple that says what _homogeneous, class C ode type is.

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  • $\begingroup$ I opine this is a matter of personal preference. I adopt the "exact form or nothing" view, so I don't see the original ODE as D'Alembert's, even though its solution space is equivalent to that of the transformed ODE's, at least modulo any extraneous solutions that may appear. It's like how I regard the ODE $y'=y$ as being separable but not linear, and its equivalent form $y'-y=0$ as linear but not separable. Also, because of this subjectiveness, I think this post should be tagged under [soft-question]. $\endgroup$ Jul 10, 2022 at 2:47
  • $\begingroup$ @AlannRosas but to solve an ODE, the very first step is to know its type, in order to apply the right method. So given the above ode, what type/name do you give it in order to apply the correct method to it? If one knows it is say D'Alembert (even after squaring), then one knows how to solve it then. So classification of an ode before solving is a very important step. For me, without this step, the solution can not even start. So this is not just an opinion issue, it is critical to writing the algorithm to solving the ode. $\endgroup$
    – Nasser
    Jul 10, 2022 at 3:02
  • $\begingroup$ Classifying an ODE certainly helps for finding solutions if general methods for the classes are known, but I don't think it's necessary for finding solutions. For instance, anyone can solve the ODE $f'=0$ without needing to classify it as linear. $\endgroup$ Jul 10, 2022 at 3:06
  • $\begingroup$ But $f'=0$ does have a classification. It is called _quadrature as shown by ode:=diff(f(x),x)=0; DEtools:-odeadvisor(ode). So knowing it is quadrature shows what method to use to solve it. Try writing an ode solver program without knowing the type of the ode. How will the algorithm know which method to apply if it does does not know the type of ode to solve? $\endgroup$
    – Nasser
    Jul 10, 2022 at 3:14
  • $\begingroup$ @AlannRosas Do you know how to solve this ode without squaring it first and then using the known D'Alembert method to solving the squared version? $\endgroup$
    – Nasser
    Jul 10, 2022 at 3:36

1 Answer 1

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As the OP asks in the comments, I’ll provide a solution without squaring the equation.

Let $z=1+x+y$. Then, $\displaystyle \frac{dz}{dx}=1+\frac{dy}{dx}$ and hence $\displaystyle \frac{dy}{dx}=\frac{dz}{dx}-1$. Thus, using the original equation we get $$\frac{dz}{dx}-1=\sqrt z\implies \frac{dz}{dx}=\sqrt z+1$$ or $$\int\frac{dz}{\sqrt z+1}=x+c \implies 2\int\frac{\sqrt z dz}{2\sqrt z(\sqrt z+1)}=x+c$$ Substituting $\sqrt z=u$ we get $$=2\int \frac{u du}{ u+1}=2(u-\ln|u+1|)+c_1$$ Thus, the general solution is $$2(u-\ln|u+1|)=x+C$$ or $$2\left(\sqrt {x+y+1}-\ln|1+\sqrt{x+y+1}|\right)=x+C.$$

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