Question

Verify that y1(x) = x is a solution of xy'' − xy' + y = 0. Use reduction of order to find a second solution y2(x) in the form of an infinite series.

A. y2 = x ln x + x^2 + x^3/ 2·2! + x^4/ 3·3! + x^5/4·4! +...
B. y2 = −1 + x ln x + x^2/ 2 + x^3/ 2·3! + x^4/ 3·4! + ...
C. y2 = ln x + x + x^2/ 2·2! + x^3/ 3·3! + x^4/ 4·4! + ...
D. y2 = ln x + x + x^2/ 2! + x^3/ 3! + x^4/ 4! + ...
E. y2 = − 1/ x + ln x + x/2 + x^2/ 2·3! + x^3/ 3·4! + ...

Conjecture an interval of definition for y2(x).
A. [0, ∞)
B. [−1, 1]
C. (−1, 1)
D. [−1, ∞)
E. (0, ∞)

Answer 1

Given $y_1(x) = x$ and differential equation is $xy'' − xy' + y = 0$ .

Now, $xy_1''-xy_1'+y_1 = x(0)-x(1)+x = 0 - x+x = 0$

Hence, $y_1(x)=x$ is solution of given differential equation.

Let the another solution is $y_2(x)$ .

we can write a second solution as:

$y_2=cy_1∫\frac{e^{∫P(x)dx}} {y_1^2}$ where $P(x) = \frac{-x}{x} = -1$ .

Hence, $y_2 = cx \int \frac{e^{-x}}{x^2}dx$ .

The closed form solution for this is ugly, so we are going to resort to a series solution.

A series solution is given by $y_2(x) =$

$cx×series \ expanssion \ of \ (\frac{e^{−x}}{x^2})=cx(\frac{1}{x^2}−\frac{1}{x}+\frac{1}{2}−\frac{x}{6}+\frac{x^2}{24}−\frac{x^3}{120}+O(x4))$ .

So. Correct option is option E = $(0,\infin)$ .