Reduction of order

$$y''+ p(x)y' + q(x)y = 0 \tag{6}$$

If we know one non-zero solution $y_{1} (x)$ somehow, then we can find the other LI solution $y_{2} (x)$
that forms a basis.

We assume that $y_{2} = v (x) y_{1}$ , where $v (x)$ is some arbitrary function.
Using the fact that $y_{1}$ and $y_{2}$ are solutions of $(6)$ , we plug it into the equation, $$y_{1}v'' + (2y_{1}' + py_{1})v' = 0$$
Dividing by $y_{1}$ and writing $v^{'} = U$ , $$U' + \left( \frac{2y_{1}'}{y_{1}} + p\right)U = 0$$
Since this is a linear equation, $$U = \frac{1}{y_{1}^2}e^{-\int pdx}$$and after substituting $v^{'} = U$ , $$v = \int \frac{1}{y_{1}^2}e^{-\int pdx}$$
Thus the general solution of $(6)$ can be written as $$y_{i}(x) = Cy_{1}(x) + Dy_{1}(x)\int \frac{1}{y_{1}^2}e^{-\int pdx}$$

Example 8

Solve $$ x y'' + (2x + 1)y' + (x + 1)y = 0$$

Solution

At $x = 0$ , the equation becomes singular, so we solve it for $x \neq 0$ .
Without loss of generality, assume $x > 0$ .
Clearly,$$y_1(x) = e^{-x}$$is a solution.
Divide the entire equation by $x$ :$$y'' + \left(2 + \frac{1}{x}\right)y' + \frac{x+1}{x}y = 0$$
Thus,$$p(x) = 2 + \frac{1}{x}$$
Using reduction of order, let$$y_2(x) = v(x) y_1(x)$$
Then$$v(x) = \int \frac{1}{y_1^2(x)} \exp\left(-\int p(x),dx\right) dx$$
Compute each part:$$y_1^2 = e^{-2x}$$

\int p (x) d x = \int (2 + \frac{1}{x}) d x = 2 x + \ln x

So,$$\exp\left(-\int p(x),dx\right)= \exp(-2x - \ln x) = \frac{e^{-2x}}{x}$$
Therefore, $$v(x) = \int \frac{1}{e^{-2x}} \cdot \frac{e^{-2x}}{x},dx = \int \frac{1}{x},dx = \ln x$$
Hence, $$ y_2(x) = e^{-x} \ln x $$
The general solution (for $x > 0$ ) is $$ y(x) = e^{-x}\left(C_1 + C_2 \ln x\right) $$

What About $x < 0$ ?

The correct general antiderivative is $$\int \frac{1}{x},dx = \ln |x|$$
So for $x < 0$ , we obtain $$v(x) = \ln |x|$$
and therefore $$ y_2(x) = e^{-x} \ln |x| $$
Thus, for $x \neq 0$ , the general solution is $$y(x) = e^{-x}\left(C_1 + C_2 \ln |x|\right)$$
Note: Since $x = 0$ is a singular point, solutions are valid separately on the intervals $(- \infty, 0)$ and $(0, \infty)$ .

How to guess a solution

Consider the homogeneous 2nd order linear differential $$y'' + p(x)y' + q(x)y = 0 \tag{7} $$then,

if $m^{2} + m p (x) + q (x) = 0$ for all $x$ then $e^{m x}$ will be a solution of $(7)$
if $m (m - 1) + m x p (x) + x^{2} q (x) = 0$ for all $x$ , then $x^{m}$ will be a solution.

$$y''+ p(x)y' + q(x)y = 0 \tag{6}$$

Solution

What About x<0?

How to guess a solution

What About $x < 0$ ?