Math Problem Statement

Find the first three nonzero terms in a Taylor polynomial approximation for the initial value problem y' = x^2 + y^2, y(0) = 1.

Solution

To solve this problem, we need to find the first three nonzero terms in the Taylor polynomial approximation for the function y(x)y(x) around x=0x = 0, given the initial condition y(0)=1y(0) = 1 and the differential equation y=x2+y2y' = x^2 + y^2.

Let's proceed step-by-step:

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

  1. Given: y(0)=1y(0) = 1.
  2. To find y(0)y'(0): y=x2+y2y' = x^2 + y^2 Substitute x=0x = 0 and y(0)=1y(0) = 1: y(0)=02+12=1y'(0) = 0^2 + 1^2 = 1

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

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

To find y(0)y''(0), we differentiate y=x2+y2y' = x^2 + y^2 with respect to xx: y=2x+2yyy'' = 2x + 2y y' Now substitute x=0x = 0, y(0)=1y(0) = 1, and y(0)=1y'(0) = 1: y(0)=20+211=2y''(0) = 2 \cdot 0 + 2 \cdot 1 \cdot 1 = 2

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

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

To find y(0)y'''(0), we need to differentiate y=2x+2yyy'' = 2x + 2y y' again: y=2+2(yy+yy)y''' = 2 + 2(y' \cdot y' + y \cdot y'') Substitute x=0x = 0, y(0)=1y(0) = 1, y(0)=1y'(0) = 1, and y(0)=2y''(0) = 2: y(0)=2+2(11+12)=2+2(1+2)=2+23=2+6=8y'''(0) = 2 + 2(1 \cdot 1 + 1 \cdot 2) = 2 + 2(1 + 2) = 2 + 2 \cdot 3 = 2 + 6 = 8

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

Step 4: Form the Taylor Polynomial Terms

The Taylor polynomial expansion for y(x)y(x) around x=0x = 0 is: y(x)y(0)+y(0)x+y(0)2x2+y(0)6x3y(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)1+1x+22x2+86x3y(x) \approx 1 + 1 \cdot x + \frac{2}{2}x^2 + \frac{8}{6}x^3 Simplify each term: y(x)1+x+x2+43x3y(x) \approx 1 + x + x^2 + \frac{4}{3}x^3

Answering the Questions

  1. I. y(0)y''(0): The answer is 2\textbf{2}, so enter "d".
  2. II. The first nonzero term: This is 1\textbf{1}, so enter "a".
  3. III. The second nonzero term: This is x\textbf{x}, so enter "b".
  4. IV. The third nonzero term: This is x2\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)y^{(4)}(0) if needed?
  2. What is the general formula for the nn-th derivative in terms of yy and xx?
  3. How does the Taylor polynomial change if y(0)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.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Taylor Series
Polynomial Approximation

Formulas

Taylor Series: y(x) ≈ y(0) + y'(0)x + (y''(0)/2)x^2 + (y'''(0)/6)x^3
Differentiation to find higher-order derivatives

Theorems

Taylor Series Expansion

Suitable Grade Level

Undergraduate (Calculus or Differential Equations level)