Science:Math Exam Resources/Courses/MATH103/April 2017/Question 08 (a)/Solution 2

We know that the coefficients of the Taylor series are given by $a_{n} = \frac{y^{(n)} (0)}{n!}$ and that the initial condition is

y (0) = 4 .

Since $y$ solves the differential equation $y^{'} + 2 y = x^{2}$ , or equivalently, $y^{'} = x^{2} - 2 y$ , we must have

y^{'} (0) = 0^{2} - 2 y (0) = 0 - 2 \cdot 4 = - 8 .

By differentiating the equation for $y^{'}$ (with respect to $x$ ), we obtain $y^{″} = 2 x - 2 y^{'}$ , which implies that

y^{″} (0) = 2 \cdot 0 - 2 y^{'} (0) = 0 - 2 \cdot - 8 = 16 .

Differentiating the equation once more, we obtain $y^{‴} = 2 - 2 y^{″}$ , so

y^{‴} (0) = 2 - 2 y^{″} (0) = 2 - 2 \cdot 16 = - 30 .

We conclude that $a_{0} = \frac{y (0)}{0!} = 4$ , $a_{1} = \frac{y^{'} (0)}{1!} = - 8$ , $a_{2} = \frac{y^{″} (0)}{2!} = 8$ , and $a_{3} = \frac{y^{‴} (0)}{3!} = - 5$ .