91Ó°ÊÓ

\ Before we start working on the problem, it's essential to understand Taylor series expansion for a function. The Taylor series expansion for a function \(f(x)\) about a point \(x = c\) is given by $$ f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f^{(3)}(c)}{3!}(x-c)^3 + \cdots $$ For our problem, we need to find the Taylor series for the function \(y(x)\) (the solution to the given differential equation) about the point \(x=0\).

\ To approximate the function using Taylor's theorem, we'll first need to compute the first few derivatives of the function. Given \(y'(x) = y(x)^2 - x^2\), we want to find: $$ y(0)=1, \quad y'(0), \quad y''(0), \quad y^{(3)}(0) $$ 1. We know \(y(0) = 1\). 2. Since \(y'(x) = y^2 - x^2\), we have \(y'(0) = y(0)^2 - 0^2 = 1^2 - 0^2 = 1 - 0 = 1\). 3. We differentiate \(y'(x)\) with respect to \(x\) to get \(y''(x)\): $$ y''(x) = \frac{d}{dx}(y^2 - x^2) = 2y(x)y'(x) - 2x $$ So, \(y''(0) = 2y(0)y'(0) - 2(0) = 2(1)(1) - 0 = 2\). 4. We differentiate \(y''(x)\) with respect to \(x\) to get \(y^{(3)}(x)\): $$ y^{(3)}(x) = \frac{d}{dx}(2yy' - 2x) = 2(y'(x)y'(x) + yy''(x) - 1) $$ Thus, \(y^{(3)}(0) = 2(y'(0)y'(0) + y(0)y''(0) - 1) = 2(1(1) + 1(2) - 1) = 4\).

\ Now that we have computed the first few derivatives and their values at \(x=0\), we can expand the function \(y(x)\) using Taylor's theorem: $$ y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y^{(3)}(0)}{3!}x^3 + \cdots $$ Plugging in the values we found in Step 2, we get: $$ y(x) = 1 + 1 \cdot x + \frac{2}{2}x^2 + \frac{4}{6}x^3 + \cdots $$ Simplifying, $$ y(x) = 1 + x + x^2 + \frac{2}{3}x^3 + \cdots $$

\ Now that we have the Taylor series for \(y(x)\), we can approximate the solution over the interval \(0 \le x \le \frac{1}{2}\) by including an adequate number of terms. For this exercise, let's keep the first three terms of the Taylor series: $$ y(x) \approx 1 + x + x^2 $$ This should provide a reasonably accurate approximation of the function \(y(x)\) over the interval \(0 \le x \le \frac{1}{2}\).