# Question:how is this ode dAlembert?

## Question:how is this ode dAlembert?

Maple 2020

I think Maple is wrong here. But may be someone could show me how it is correct?

Maple says this ode (below) is of type d'Alembert. But I am not able to show this. It is impossible for me to put this ode in _dAlembert. form. So I gave up.

The challenge then is to put the following first order ODE in the above form to show it is dAlembert.

I could not do it. I worked on this by hand and not possible to get the ODE in the above form. Could someone show this?

```restart;
ode:=3*x^2*y(x)^3+y(x)^4+(3*x^3*y(x)^2+4*x*y(x)^3+y(x)^4)*diff(y(x),x) = 0;

The first thing I do when I want to show this, is to solve for y(x) from the ode. Since I can't use solve on an ode, I start by replacing all the diff(y(x),x) with say p. Then now solve for y(x). If it is dAlembert, then it should give expression that be put in the form    y(x)=x*f(p) + g(p). Notice that the functions f(p) and g(p) are functions of p only and not of x. This is important.  And f(p) is multiplied by linear term and not x^(3/2) or x^(1/2), etc... The term multiplying f(p) has to be linear in x.

```ode:=subs(diff(y(x),x)=p,ode):
sol:=[PDEtools:-Solve(ode,y(x))];
```

Looking at second and third solutions. None of them is dAlembert.  This can be shown by either simplyfing it with assumptions, where not possible to obtain the needed form, or by simply replacing p back with diff(y(x),x) and asking advisor for the type of the resulting ode

```DEtools:-odeadvisor( subs(p=diff(y(x),x),sol[2]));
```

So none is d'Alembert.

Question is: Could someone may be proof that this ode is d'Alembert? By putting it in the form   y(x)=x*f(p)+g(p)? Or is advisor is wrong here?

ps. I tried infolevel[DEtools:-odeadvisor]:=4 to try to trace it, but it does not work.

pps. I worked this out by hand, and I get

y(x)= x^(3/2)*f(p)  where f(p) = sqrt(-12 p^2)+sqrt(12*p)

And this is not d'Alembert.

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