The given differential equation is:

\[
\frac{1}{3}y' = \left(\frac{1}{3}x'\right)^2 + 4\left(\frac{1}{3}x'\right) - 5
\]

### Step-by-Step Simplification:

1. Multiply both sides by 3 to eliminate the \(\frac{1}{3}\) coefficient:
   \[
   y' = 3 \left[ \left(\frac{1}{3}x'\right)^2 + 4\left(\frac{1}{3}x'\right) - 5 \right]
   \]

2. Simplify each term inside the brackets:
   \[
   \left(\frac{1}{3}x'\right)^2 = \frac{1}{9}(x')^2
   \]
   \[
   4\left(\frac{1}{3}x'\right) = \frac{4}{3}x'
   \]

3. Now substitute these into the equation:
   \[
   y' = 3 \left[\frac{1}{9}(x')^2 + \frac{4}{3}x' - 5\right]
   \]

4. Distribute the factor of 3:
   \[
   y' = \frac{3}{9}(x')^2 + 3 \times \frac{4}{3}x' - 3 \times 5
   \]
   \[
   y' = \frac{1}{3}(x')^2 + 4x' - 15
   \]

Thus, the simplified form of the equation is:

\[
y' = \frac{1}{3}(x')^2 + 4x' - 15
\]

Let me know if you'd like further explanation or details on how to proceed with solving this.  

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Here are five related questions:
1. How do you solve a first-order linear differential equation?
2. What are the common methods to solve a quadratic differential equation?
3. Can this equation be interpreted geometrically in terms of slopes?
4. How would the solution change if the right-hand side included a higher degree polynomial?
5. What is the role of initial conditions when solving this kind of differential equation?

**Tip**: Always check if multiplying through by a constant simplifies coefficients, making the equation easier to manage.

This article provides a step-by-step solution to simplifying the differential equation 1/3 y' = (1/3 x')² + 4(1/3 x') - 5. The solution involves simplifying quadratic expressions and differential equations, ideal for advanced high school students (Grades 11-12).

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