## Solving the Differential Equation (y-x)y' = y-x+8 with the given function y=x+4(x+2)^(1/2)

This article explores the solution to the differential equation **(y-x)y' = y-x+8** with the given function **y=x+4(x+2)^(1/2)**. We will verify if the given function is indeed a solution to the differential equation.

### 1. Finding y'

Firstly, we need to find the derivative of the given function **y=x+4(x+2)^(1/2)**.

Using the power rule and chain rule, we get:
**y' = 1 + 2(x+2)^(-1/2)**

### 2. Substituting y and y' into the Differential Equation

Now, we substitute **y** and **y'** into the original differential equation:

**(y-x)y' = y-x+8**

**[x+4(x+2)^(1/2) - x] * [1 + 2(x+2)^(-1/2)] = [x+4(x+2)^(1/2) - x] + 8**

Simplifying the equation:

**4(x+2)^(1/2) * [1 + 2(x+2)^(-1/2)] = 4(x+2)^(1/2) + 8**

**4(x+2)^(1/2) + 8 = 4(x+2)^(1/2) + 8**

### 3. Conclusion

As we can see, both sides of the equation are equal. This verifies that the given function **y=x+4(x+2)^(1/2)** is indeed a **solution** to the differential equation **(y-x)y' = y-x+8**.

In conclusion, we have successfully verified that the given function satisfies the given differential equation.