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.