To find the first three nonzero terms of the Taylor series solution for the initial value problem $4x'' + 8tx = 0$, with $x(0) = 1$ and $x'(0) = 0$, let's proceed step-by-step.

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

The Taylor series for $x(t)$ is: $x(t) = x(0) + x'(0)t + \frac{x''(0)}{2!}t^2 + \frac{x^{(3)}(0)}{3!}t^3 + \cdots$ Given $x(0) = 1$ and $x'(0) = 0$, we already know: $x(t) = 1 + 0 \cdot t + \frac{x''(0)}{2}t^2 + \frac{x^{(3)}(0)}{6}t^3 + \cdots$

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Step 2: Compute $x''(0)$ and higher derivatives

The given differential equation is: $4x'' + 8tx = 0$ Simplify: $x'' = -2tx$

(i) First derivative, $x'(t)$:

From the problem, $x'(0) = 0$.

(ii) Second derivative, $x''(t)$:

Using $x'' = -2tx$ and substituting $t = 0$ and $x(0) = 1$: $x''(0) = -2 \cdot 0 \cdot 1 = 0$

(iii) Third derivative, $x^{(3)}(t)$:

Differentiate $x'' = -2tx$ with respect to $t$: $x''' = -2x - 2tx'$ Substitute $t = 0$, $x(0) = 1$, and $x'(0) = 0$: $x'''(0) = -2 \cdot 1 - 2 \cdot 0 \cdot 0 = -2$

(iv) Fourth derivative, $x^{(4)}(t)$:

Differentiate $x''' = -2x - 2tx'$: $x^{(4)} = -2x' - 2(x' + tx'')$ Substitute $t = 0$, $x(0) = 1$, $x'(0) = 0$, and $x''(0) = 0$: $x^{(4)}(0) = -2 \cdot 0 - 2(0 + 0) = 0$

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Step 3: Taylor series expansion

Now substitute the computed derivatives into the Taylor series: $x(t) = x(0) + x'(0)t + \frac{x''(0)}{2!}t^2 + \frac{x^{(3)}(0)}{3!}t^3 + \frac{x^{(4)}(0)}{4!}t^4 + \cdots$ Substitute $x(0) = 1$, $x'(0) = 0$, $x''(0) = 0$, and $x^{(3)}(0) = -2$: $x(t) = 1 + 0 \cdot t + \frac{0}{2!}t^2 + \frac{-2}{6}t^3 + \frac{0}{24}t^4 + \cdots$ Simplify: $x(t) = 1 - \frac{1}{3}t^3 + \cdots$

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Final Answer:

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

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Questions:

1. Would you like to compute additional terms in the Taylor series?
2. Do you need help visualizing the Taylor polynomial or the original solution?
3. Should we discuss the convergence of this series?
4. Would you like to verify the solution using an alternate method?
5. Do you need an explanation on deriving higher-order derivatives for Taylor series?

Tip: Always check initial conditions and substitute step-by-step when working with Taylor series expansions for differential equations.