To solve this problem, we need to find the first three nonzero terms in the Taylor polynomial approximation for the function $y(x)$ around $x = 0$, given the initial condition $y(0) = 1$ and the differential equation $y' = x^2 + y^2$.

Let's proceed step-by-step:

Step 1: Initial Values and Derivatives at $x = 0$

1. Given: $y(0) = 1$.
2. To find $y'(0)$: $y' = x^2 + y^2$ Substitute $x = 0$ and $y(0) = 1$: $y'(0) = 0^2 + 1^2 = 1$

So, $y'(0) = 1$.

Step 2: Compute $y''(0)$

To find $y''(0)$, we differentiate $y' = x^2 + y^2$ with respect to $x$: $y'' = 2x + 2y y'$ Now substitute $x = 0$, $y(0) = 1$, and $y'(0) = 1$: $y''(0) = 2 \cdot 0 + 2 \cdot 1 \cdot 1 = 2$

So, $y''(0) = 2$.

Step 3: Compute $y'''(0)$

To find $y'''(0)$, we need to differentiate $y'' = 2x + 2y y'$ again: $y''' = 2 + 2(y' \cdot y' + y \cdot y'')$ Substitute $x = 0$, $y(0) = 1$, $y'(0) = 1$, and $y''(0) = 2$: $y'''(0) = 2 + 2(1 \cdot 1 + 1 \cdot 2) = 2 + 2(1 + 2) = 2 + 2 \cdot 3 = 2 + 6 = 8$

So, $y'''(0) = 8$.

Step 4: Form the Taylor Polynomial Terms

The Taylor polynomial expansion for $y(x)$ around $x = 0$ is: $y(x) \approx y(0) + y'(0)x + \frac{y''(0)}{2}x^2 + \frac{y'''(0)}{6}x^3$

Now substitute the values we found: $y(x) \approx 1 + 1 \cdot x + \frac{2}{2}x^2 + \frac{8}{6}x^3$ Simplify each term: $y(x) \approx 1 + x + x^2 + \frac{4}{3}x^3$

Answering the Questions

1. I. $y''(0)$: The answer is $\textbf{2}$, so enter "d".
2. II. The first nonzero term: This is $\textbf{1}$, so enter "a".
3. III. The second nonzero term: This is $\textbf{x}$, so enter "b".
4. IV. The third nonzero term: This is $\textbf{x}^2$, so enter "d".

Would you like further details on each step?

Here are five additional questions for practice:

1. How would you find $y^{(4)}(0)$ if needed?
2. What is the general formula for the $n$-th derivative in terms of $y$ and $x$?
3. How does the Taylor polynomial change if $y(0)$ had a different initial value?
4. Can you derive the Taylor series from the differential equation directly without calculating individual derivatives?
5. How would the Taylor series change if the differential equation were different?

Tip: Remember to substitute initial values after differentiating instead of before.