To solve the given problem using the fx-991CW calculator, follow these steps. The task is to calculate $y(0.1)$ and $y(0.2)$ using the Taylor series approximation up to the third derivative.

Given:

• Differential equation:
  $\frac{dy}{dx} = x^2y - 1, \quad y(0) = 1$
• Taylor series expansion formula:
  $y(x) = y(x_0) + \frac{x - x_0}{1!}y'(x_0) + \frac{(x - x_0)^2}{2!}y''(x_0) + \frac{(x - x_0)^3}{3!}y'''(x_0) + \cdots$

Steps to Calculate Using fx-991CW

1. Store Initial Values:

   • Set $x_0 = 0$ and $y(0) = 1$.
   • You know:
     $y'(0) = -1, \quad y''(0) = 0, \quad y'''(0) = 2$

2. Taylor Series Formula Input: Enter the series expansion formula up to the third derivative: $y(x) = y(0) + x \cdot y'(0) + \frac{x^2}{2} \cdot y''(0) + \frac{x^3}{6} \cdot y'''(0)$ Substituting $y(0) = 1, y'(0) = -1, y''(0) = 0, y'''(0) = 2$, we get: $y(x) = 1 - x + \frac{x^3}{3}$

3. Use Calculator to Compute $y(0.1)$ and $y(0.2)$:

   • Press the MODE button and select POLYNOMIAL or TABLE mode.
   • Enter the function $y(x) = 1 - x + \frac{x^3}{3}$.
   • Use the calculator's table or direct evaluation to compute for $x = 0.1$ and $x = 0.2$.

4. Results:

   • For $x = 0.1$:
     $y(0.1) = 1 - 0.1 + \frac{(0.1)^3}{3} = 0.9003$
   • For $x = 0.2$:
     $y(0.2) = 1 - 0.2 + \frac{(0.2)^3}{3} = 0.8023$

5. Verify Decimal Precision: The calculator supports up to 5 decimal places. Ensure mode settings reflect this for accurate results.

Explanation of Derivatives

The derivatives are computed as follows:

1. $y' = x^2y - 1$, at $x = 0, y = 1$:
   $y'(0) = (0^2)(1) - 1 = -1$
2. $y'' = 2xy + x^2y'$, at $x = 0, y = 1, y'(0) = -1$:
   $y''(0) = 2(0)(1) + (0^2)(-1) = 0$
3. $y''' = 2y + 4xy' + x^2y''$, at $x = 0, y = 1, y'(0) = -1, y''(0) = 0$:
   $y'''(0) = 2(1) + 4(0)(-1) + (0^2)(0) = 2$

Would you like further clarifications or assistance?

Related Questions:

1. How to apply Taylor series to solve higher-order ODEs?
2. How does the fx-991CW handle factorial terms for Taylor expansion?
3. How can this method extend to approximate values at more distant points (e.g., $x = 0.5$)?
4. What are the limitations of the Taylor series method?
5. How can the calculator be used to verify derivatives directly?

Tip:

Always store intermediate results in the calculator's memory to reduce rounding errors during computation!