function nlinbvp4c
solinit = bvpinit(linspace(0,1,5),[0 0]);
sol = bvp4c(@twoode,@twobc,solinit);
function dydx = twoode(x,y)
dydx = [ y(2); abs(y(1))*(1+y(2)^2)^0.75 ];
function res = twobc(ya,yb)
res = [ ya(2); yb(2) - 1 ];
Not sure, but when solving for y'', you may also take the negative square root of the right-hand side:
y''=-abs(y(1))*(1+y(2)^2)^0.75
Thus there might be two solutions for your problem.
Best wishes
Torsten.
