Use power series to solve the differential equation..

Question

Quizplus · Accepted Answer

To solve the differential equation using power series, we need to express the solution as a power series expansion. Let y be the solution to the differential equation:
$$y'' + 4xy' + y = 0$$
Let y be represented as a power series:
$$y(x) = \sum_{n=0}^{\infty} a_n x^n$$
Then differentiate twice with respect to x to obtain:
$$y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}$$
$$y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}$$
Substituting these expansions of y, y', and y'' into the differential equation, and equating the coefficients of like powers of x, we obtain:
$$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + 4x\sum_{n=1}^{\infty} na_n x^{n-1} + \sum_{n=0}^{\infty} a_n x^n = 0$$
Rearranging and simplifying, we obtain:
$$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + 4na_n + a_n]x^n = 0$$
Since x can be any value, the coefficient of each x^n term must be equal to zero, giving us the recurrence relation:
$$(n+2)(n+1)a_{n+2} + (4n+1)a_n = 0$$
We can use this recurrence relation to obtain the coefficients of the power series for y. The first few coefficients are:
$$a_0 = 1$$
$$a_1 = 0$$
$$a_2 = -\frac{1}{2\cdot1(2+1)}a_0 = -\frac{1}{6}$$
$$a_3 = -\frac{1}{3\cdot2(3+1)}a_2 = \frac{1}{90}$$
$$a_4 = -\frac{1}{4\cdot3(4+1)}a_2 - \frac{1}{4\cdot3(4+1)}a_0 =\frac{1}{360} - \frac{1}{60} = -\frac{1}{180} $$
Thus, the power series expansion for y is:
$$y(x) = 1 - \frac{1}{6}x^2 + \frac{1}{90}x^4 - \frac{1}{180}x^6 + \cdots$$
which converges for all values of x. Therefore, the solution to the differential equation is given by:
$$y(x) = 1 + \sum_{n=2}^{\infty} (-1)^n \frac{x^{2n}}{(2n)! (n-1)!}$$
which is equivalent to choice C.

Use Power Series to Solve the Differential Equation