Problem Statement
Consider the differential equation $\frac{dy}{dx} = 2x + y^2$ with initial condition $y(0) = 1$.
(a) Find the values of $\frac{dy}{dx}$, $\frac{d^2 y}{dx^2}$, and $\frac{d^3 y}{dx^3}$ at $x = 0$.
(b) Hence, find the Maclaurin series for $y$ up to and including the term in $x^3$.
Verified Solution & Marking Scheme
Evaluate first derivative at x = 0
Given $\frac{dy}{dx} = 2x + y^2$.
At $x = 0, y = 1$:
$$\left.\frac{dy}{dx}\right|_{x=0} = 2(0) + (1)^2 = 1$$
Differentiate implicitly to find second derivative
$$\frac{d^2 y}{dx^2} = \frac{d}{dx}(2x + y^2) = 2 + 2y \frac{dy}{dx}$$
At $x = 0, y = 1, y' = 1$:
$$\left.\frac{d^2 y}{dx^2}\right|_{x=0} = 2 + 2(1)(1) = 2 + 2 = 4$$
Differentiate again to find third derivative
Using product rule on $2y \frac{dy}{dx}$:
$$\frac{d^3 y}{dx^3} = \frac{d}{dx}\left(2 + 2y \frac{dy}{dx}\right) = 0 + 2\left(\frac{dy}{dx}\right)^2 + 2y \frac{d^2 y}{dx^2}$$
At $x = 0$:
$$\left.\frac{d^3 y}{dx^3}\right|_{x=0} = 2(1)^2 + 2(1)(4) = 2 + 8 = 10$$
Construct Maclaurin series expansion
Formula from IB Formula Booklet:
$$y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$$
Substitute values:
$$y(x) = 1 + 1x + \frac{4}{2}x^2 + \frac{10}{6}x^3 = 1 + x + 2x^2 + \frac{5}{3}x^3$$