Let us dsolve that Cauchy problem with 12.3 on Windows 10 Pro:
ClearAll[w, z]; sol = DSolve[{w'[z] == -1/2 - Sqrt[1/4 - 3*w[z]], w[1] == -1}, w[z], z]
{{w[z] -> 1/12 (-2 ProductLog[-((1 - Sqrt[13]) E^(2 + Sqrt[13] - 3 z))] - ProductLog[-((1 - Sqrt[13]) E^(2 + Sqrt[13] - 3 z))]^2)}, {w[ z] -> 1/12 (-2 ProductLog[(-1 + Sqrt[13]) E^( 2 + Sqrt[13] - 3 z)] - ProductLog[(-1 + Sqrt[13]) E^(2 + Sqrt[13] - 3 z)]^2)}}
and a warning "Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information". Both results are not correct as
Plot[w[z] /. sol[[1]], {z, 1, 2}]
Plot[w[z] /. sol[[2]], {z, 1, 2}]
show: w'[-1]==-1/2-Sqrt[13/4]
, making use of {w'[z] == -1/2 - Sqrt[1/4 - 3*w[z]], w[1] == -1}
, is negative, whereas we see a positive slope at x==1
in the plots. What is the reason of that incorrect result and how to fix it?
w'[-1] -> -1 - 1/2 ProductLog[(-1 + Sqrt[13]) E^(5 + Sqrt[13])]
. Nevertheless, from the first glance, it is not clear what reason brought you to the conclusion that the resultw'[-1]==-1/2-Sqrt[13/4]
is wrong? Explain please. $\endgroup$DSolve[{w'[z] == -1/2 - Sqrt[1/4 - 3*w[z]], w[1] == -1}, w[z], z]
are wrong since their plots have a positive slop atx==1
. $\endgroup${w'[z] == -1/2 - Sqrt[1/4 - 3*w[z]], w[1] == -1}
to obtainw'[-1]==-1/2-Sqrt[13/4]
. $\endgroup$DSolve[w'[z]^2+w'[z]+3*w[z]==0,w,z]
$\endgroup$