why do i receive error when do 'Implicit' solution true?

syms  y(x) 
eqn=diff(y,x)==3*(x*y)^(1/2); 
dsolve(eqn,'Implicit',true) 

when i run this code

ans =

y(x) == 0

(y(x)^(3/2)*(C1 + 27*y(x)^(1/2)))^(2/3)/y(x) == (9*x)/((3^(1/2)*1i)/2 + 1/2)^2

(y(x)^(3/2)*(C1 + 27*y(x)^(1/2)))^(2/3)/y(x) == (9*x)/((3^(1/2)*1i)/2 - 1/2)^2

(y(x)^(3/2)*(C1 + 27*y(x)^(1/2)))^(2/3)/y(x) == 9*x

programme prints on the screen this but implicit solution of this diferential equation is x^(3/2)-sqrt(y)=C

if someone can help me it would be great, thanks!!!

Answer 1

[Walter Roberson]

sol = dsolve(eqn, 'implicit', true)
arrayfun(@(X)isolate(X, y), sol)

You will see that they are different ways of writing it. There might potentially be theoretic differences if x or y(x) are negative or complex valued

[sedat]

thanks for the answer but it still doesn't print x^(3/2)-sqrt(y)=C which is implicit solution of the problem.

[Walter Roberson]

syms  y(x) 
eqn=diff(y,x)==3*(x*y)^(1/2); 
sol = dsolve(eqn, 'implicit', true)

sol = 

sol1 = arrayfun(@(X)isolate(X, y), sol);
arrayfun(@(X) sqrt(lhs(X)) == simplify(sqrt(rhs(X)), 'ignoreanalytic', true), sol1)

ans = 

Notice that analytic constraints had to be ignored. You would have to prove that your version is valid for negative x.