y' = sqrt(x)y Solve the differential equation

An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows (dy)/(dx)= f(x,y) .
It can also be in a form of N(y) dy= M(x) dx as variable separable differential equation..
 To be able to set-up the problem as N(y) dy= M(x) dx , we let y' = (dy)/(dx).
 The problem: y'=sqrt(x)y becomes:
(dy)/(dx)=sqrt(x)y
Rearrange by cross-multiplication, we get:
(dy)/y=sqrt(x)dx
Apply direct integration on both sides: int (dy)/y=int sqrt(x)dx  to solve for the general solution of a differential equation.
 
For the left side, we applying basic integration formula for logarithm:
int(dy)/y= ln|y|
For the right side, we apply the Law of Exponent: sqrt(x)=x^(1/2) then follow the Power Rule of integration: int x^n dx = x^(n+1)/(n+1)+C
int sqrt(x)* dx= int x^(1/2)* dx
                   = x^(1/2+1)/(1/2+1)+C
                  = x^(3/2)/(3/2)+C
                  = x^(3/2)*(2/3)+C
                  = (2x^(3/2))/3+C
Combining the results from both sides, we get the general solution of the differential equation as:
ln|y|=(2x^(3/2))/3+C
or
y =e^(((2x^(3/2))/3+C))

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