Solving ODE through reduction of order

Question

Verify that $y_1(x) = x$ is a solution of $xy''-xy'+y = 0$. Use reduction of order to find a second solution $y_2(x)$ in the form of an infinite series.

Conjecture an interval of definition for $y_2(x)$.

Any ideas?

Answer 1

Theory - Reduction of Order

When we have a homogeneous second-order linear ordinary differential equation of the form:

$$\tag 1 a_2(x) y'' + a_1(x) y' + a_0(x) y = 0,$$

we can write it in standard form by dividing through by $a_2(x)$, while noting problems areas, as:

$$\tag 2 y'' + P(x) y' + Q(x) y = 0$$

$P(x)$ and $Q(x)$ are continuous on some interval $I$. If we know a solution $y_1(x)$ of $(2)$ on $I$, and that $y_1(x) \ne 0$ for every $x$ in I, then we can write a second solution as:

$$\tag 3 y_2(x) = y_1(x)\int \dfrac{e^{\int P(w) dw}}{y^2_1(x)}\, dx$$

Problem

• $(a)$ Verify that $y_1(x) = x$ is a solution of $xy''-xy'+y = 0$.
• $(b)$ Use reduction of order to find a second solution $y_2(x)$ in the form of an infinite series.
• $(c)$ Conjecture an interval of definition for $y_2(x)$.

Part a

We are given:

$\tag 4 xy''-xy'+y = 0$ and $y_1(x) = x$ and asked to verify that it is a solution.

So we have: $y_1(x) = x, y'_1(x) = 1, y''_2(x) = 0$ and substituting into $(4)$ yields:

$$xy''-x y'+ y = x(0) -x (1) + x = 0$$

Part b

Using the ROO $(3)$, we can write a second solution as:

$\displaystyle y_2 = y_1\int \dfrac{e^{\int P(w) dw}}{y^2_1}\, dx = x \int \dfrac{e^{\int -dw}}{x^2}\, dx = x \int \dfrac{e^{(-x + c)}}{x^2}\, dx = x \int \dfrac{c e^{-x}}{x^2}\, dx = cx \int \dfrac{e^{-x}}{x^2}\, dx$

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

A series solution (do you know how to do a Taylor series expansion for the expression) is given by:

$$\displaystyle cx \times \text{series exp of} \left(\frac{e^{-x}}{x^2}\right) = cx \left(\frac{1}{x^2} - \frac{ 1}{x} + \frac{1}{2} -\frac{x}{6} + \frac{x^2}{24} - \frac{x^3}{120} + O(x^4)\right)$$

This can be compactly written as:

$$\tag 5 y_2(x) = cx \sum_{n=-2}^\infty \frac{(-1)^n x^n}{(n+2)!}$$

You should very that $(5)$ is indeed a solution to the original DEQ.

Part c

You should be able to handle this.

Lastly, it is worth noting that the general solution is given by:

$$y(x) = c_1 y_1(x) + c_2 y_2(x) = c_1 x + c_2 \left(cx \sum_{n=-2}^\infty \frac{(-1)^n x^n}{(n+2)!}\right)$$

over the appropriate interval.