To solve the initial value problem and determine the first three nonzero terms in the Taylor series expansion for $x(t)$, let us carefully go step by step:

The given equation is:

$2x'' + 9t x = 0, \quad x(0) = 1, \quad x'(0) = 0.$

We are tasked with finding the Taylor expansion up to three nonzero terms.

Step 1: Write the Taylor series form for $x(t)$:

$x(t) = x(0) + x'(0)t + \frac{x''(0)}{2!}t^2 + \frac{x'''(0)}{3!}t^3 + \cdots$

Step 2: Use initial conditions:

From $x(0) = 1$ and $x'(0) = 0$, we have the first two terms: $x(t) = 1 + 0t + \frac{x''(0)}{2!}t^2 + \frac{x'''(0)}{3!}t^3 + \cdots$

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

From the differential equation $2x'' + 9t x = 0$, substitute $t = 0$: $2x''(0) + 9(0)x(0) = 0 \implies x''(0) = 0.$

Thus, the Taylor expansion becomes: $x(t) = 1 + 0t + 0t^2 + \frac{x'''(0)}{3!}t^3 + \cdots$

Step 4: Compute $x'''(0)$:

Differentiate the equation $2x'' + 9t x = 0$ with respect to $t$: $2x''' + 9(x + tx') = 0.$

At $t = 0$, substitute $x(0) = 1$ and $x'(0) = 0$: $2x'''(0) + 9(1 + 0) = 0 \implies 2x'''(0) + 9 = 0 \implies x'''(0) = -\frac{9}{2}.$

Now the Taylor expansion becomes: $x(t) = 1 + 0t + 0t^2 + \frac{-\frac{9}{2}}{6}t^3 + \cdots = 1 - \frac{3}{4}t^3 + \cdots$

Final Result:

The first three nonzero terms of the Taylor approximation are: $x(t) = 1 - \frac{3}{4}t^3 + \cdots$

Would you like a deeper explanation of any step or related expansions?

Here are five related questions for further understanding:

1. How is the Taylor series derived for a general differential equation?
2. Why do we compute derivatives up to the required terms for Taylor expansions?
3. Can you explore how higher-order terms would change the approximation?
4. What are the practical uses of Taylor approximations in solving differential equations?
5. How does the convergence of a Taylor series affect its accuracy in approximating solutions?

Tip: Always cross-verify computed derivatives and coefficients when deriving Taylor expansions to ensure accuracy.