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