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)`

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