sqrt(x) + sqrt(y)y' = 0 , y(1) = 9 Find the particular solution that satisfies the initial condition

For the given problem: sqrt(x)+sqrt(y)y' =0, we may rearrange this to
sqrt(y)y' = -sqrt(x)
Recall that y' is denoted as (dy)/(dx) then it becomes:
sqrt(y)(dy)/(dx) = -sqrt(x)
Apply the variable separable differential equation in a form of f(y) dy = g(x) dx .
sqrt(y)(dy) = -sqrt(x)dx
Apply direct integration using the Power Rule: int u^n du = u^(n+1)/(n+1) .
Note: sqrt(x) = x^(1/2) and sqrt(y) = y^(1/2) .
int sqrt(y)(dy) = int -sqrt(x)dx
int y^(1/2) (dy) = int -x^(1/2)dx
y^(1/2+1)/(1/2+1)= -x^(1/2+1)/(1/2+1) +C
y^(3/2)/(3/2) = -x^(3/2)/(3/2)+C
y^(3/2)*2/3 = -x^(3/2)*2/3+C
2/3y^(3/2) = -2/3x^(3/2)+C
 
The general solution of the differential equation is 2/3y^(3/2)= -2/3x^(3/2)+C .
Using the given initial condition y(1)=9 , we plug-in x=1 and y=9  to solve for C:
2/3(9)^(3/2)= -2/3*1^(3/2)+C
2/3*27=-2/3*1+C
18=-2/3+C
C = 18+2/3
C = 56/3
So,
2/3y^(3/2)= -2/3x^(3/2)+56/3
y^(3/2)=(3/2)(-2/3x^(3/2)+56/3)
y^(3/2)=-x^(3/2)+28
(y^(3/2))^(2/3)=(-x^(3/2)+28)^(2/3)
y =(-x^(3/2)+28)^(2/3)
or
y = root(3)((-x^(3/2)+28)^2)