Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 6

An ordinary differential equation (ODE) is differential equation for the derivative of a function of one variable. When an ODE is in a form of y'=f(x,y) , this is just a first order ordinary differential equation.
The given problem: y' = -sqrt(x)/(4y) is in a form of y'=f(x,y) .
To evaluate this, we may express y' as (dy)/(dx) .
The problem becomes:
(dy)/(dx)= -sqrt(x)/(4y)
We may apply the variable separable differential equation: N(y) dy = M(x) d x.
Cross-multiply dx to the right side:
dy= -sqrt(x)/(4y)dx
Cross-multiply 4y to the left side:
4ydy= -sqrt(x)dx
Apply direct integration on both sides:
int 4ydy= int -sqrt(x)dx
Apply basic integration property: int c*f(x)dx = c int f(x) dx on both sides:
4 int ydy= (-1) int sqrt(x)dx
For the left side, we apply the Power Rule for integration: int u^n du= u^(n+1)/(n+1)+C .
4 int y dy = 4*y^(1+1)/(1+1)
= 4*y^2/2
= 2y^2

For the right side, we apply Law of Exponent: sqrt(x)= x^(1/2) before applying the Power Rule for integration: int u^n du= u^(n+1)/(n+1)+C .
(-1) int x^(1/2)dx =(-1) 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, we get the general solution for differential equation:
2y^2= -(2x^(3/2))/3+C
We may simplify it as:
(1/2)[2y^2]= (1/2)[-(2x^(3/2))/3+C]
y^2= -(2x^(3/2))/6+C/2
y= +-sqrt(-(2x^(3/2))/6+C/2)
y= +-sqrt(-(x^(3/2))/3+C/2)