# [Maxima] Fwd: How to get a reduced solution?

Fri Nov 12 11:20:46 CST 2010

```I think this solved my problem partly.

But I would like a output like that
http://www.wolframalpha.com/input/?i=y''+%3D+exp(y)
which is exactly the solution Mathematica gives.

Anyway, thank you very much.

2010/11/12 Jaime Villate <villate at fe.up.pt>

On Thu, 2010-11-11 at 14:47 -0800, Edwin Woollett wrote:
> > On Nov. 6, 2010, linwaytin wrote:
> > -----------------------------
> > > I want to solve a differential equation y'' = e^y.
> > > It seems Maxima got the solution but in the form x=f(y). I would like
> it
> > > in
> > > the form y=f(x), how to get it?
> > > By the way, I tried Mathematica and it gave me directly the form
> y=f(x).
> > > Can
> > > Maxima do that automatically?
> > -----------------------
> > I agree that Maxima's ode2 function returns only two implicit
> > solutions for y as a function of x (I do this two ways below).
> Hi Edwin,
> I think that linwaytin still wants some further simplification to get
> the solution as he wants:
>
> (%i1) display2d: false\$
>
> (%i2) de : 'diff(y,x,2) = exp (y)\$
>
> (%i3) ode2 (de,y,x);
>
> Is  %k1  positive or negative?
>
> p;
> (%o3) [-log((sqrt(%e^y+%k1)-sqrt(%k1))/(sqrt(%e^y+%k1)+sqrt(%k1)))
>         /(sqrt(2)*sqrt(%k1))
>         = x+%k2,
>       log((sqrt(%e^y+%k1)-sqrt(%k1))/(sqrt(%e^y+%k1)+sqrt(%k1)))
>        /(sqrt(2)*sqrt(%k1))
>         = x+%k2]
> (%i4) solve(%o3[1],y);
>
> Is  %k1*(%e^(sqrt(%k1)*(-sqrt(2)*x-sqrt(2)*%k2))-1)  positive, negative,
> or zero?
>
> p;
> (%o4) [sqrt(%e^y+%k1) = -sqrt(%k1)*(%e^(sqrt(%k1)*(-sqrt(2)*x-sqrt(2)*%
> k2))+1)
>                      /(%e^(sqrt(%k1)*(-sqrt(2)*x-sqrt(2)*%k2))-1)]
> (%i5) solve(%^2,y);
>
> (%o5) [y = log(4*%k1*%e^(-sqrt(2)*sqrt(%k1)*x-sqrt(2)*sqrt(%k1)*%k2)
>                /(%e^(-2^(3/2)*sqrt(%k1)*x-2^(3/2)*sqrt(%k1)*%k2)
>                 -2*%e^(-sqrt(2)*sqrt(%k1)*x-sqrt(2)*sqrt(%k1)*%k2)+1))]
> (%i6) solve(%o3[2],y);
>
> Is  %k1*(%e^(sqrt(%k1)*(sqrt(2)*x+sqrt(2)*%k2))-1)  positive, negative,
> or zero?
>
> p;
> (%o6) [sqrt(%e^y+%k1) = -sqrt(%k1)*(%e^(sqrt(%k1)*(sqrt(2)*x+sqrt(2)*%
> k2))+1)
>                      /(%e^(sqrt(%k1)*(sqrt(2)*x+sqrt(2)*%k2))-1)]
> (%i7) solve(%^2,y);
>
> (%o7) [y = log(4*%k1*%e^(sqrt(2)*sqrt(%k1)*x+sqrt(2)*sqrt(%k1)*%k2)
>                /(%e^(2^(3/2)*sqrt(%k1)*x+2^(3/2)*sqrt(%k1)*%k2)
>                 -2*%e^(sqrt(2)*sqrt(%k1)*x+sqrt(2)*sqrt(%k1)*%k2)+1))]
>
> Cheers,
> Jaime
>
>
>
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