second analytical solution of implicit equation

When I do the command

F = M*exp(-M^2/2)==(2.72498/r)^2*exp(1.5-2*(2.72498/r));

solve(F, M)

it returns only 1 analytical solution of the two. Is there a way to get the expression for the second solution? When the function is plotted implicitly you can see that it goes like the figure.

How to get to an expression for the other branches? I succeeded in plotting the braches seperatly, but I need the expression for further calculation and plotting of other values containing these M and r, composed of solutions of the different branches.

[Star Strider]

It returns two solutions in r, and one solution in M —

syms M r 
F = M*exp(-M^2/2)==(2.72498/r)^2*exp(1.5-2*(2.72498/r));
M_sol = solve(F, M) 

M_sol = 

r_sol = solve(F, r) 

r_sol = 

.

[renee]

Indeed, but my question is: how to get to the second solution in M which it clearly has.

Answer 1

[Sulaymon Eshkabilov]

This is the same as the solution proposed by Star Strider:

syms M r 
EQN = M*exp(-M^2/2)==(2.72498/r)^2*exp(1.5-2*(2.72498/r));
Solution_M =solve(EQN,M,'IgnoreAnalyticConstraints',true)

Solution_M = 

Solution_M =solve(EQN, r,'IgnoreAnalyticConstraints',true)

Solution_M = 

Answer 2

[Star Strider]

I believe you are plotting ‘M’ as a funciton of ‘r’ however it likely does not have an analytic solution. In the lambertw argument, it is a function of and more generally, , however I doubt that it is possible to factor it further. A numeric solution is the only option.

syms M r 
F = M*exp(-M^2/2)==(2.72498/r)^2*exp(1.5-2*(2.72498/r));
Fc(r,M) = M*exp(-M^2/2) - ((2.72498/r)^2*exp(1.5-2*(2.72498/r)));
M_sol = solve(F, M) 

M_sol = 

Mr_sol = solve(M_sol,r)

 
Mr_sol =
 
Empty sym: 0-by-1
 

r_sol = solve(F, r) 

r_sol = 

figure
fsurf(Fc, [0 5  0.01 5])
colormap(turbo)

figure
fimplicit(Fc, [0 5  0.01 5])
colormap(turbo)
xlabel('r')
ylabel('M')

.