Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Purpose

Before delving into the actual mechanism of doing this method, it’s important to think of the big picture.

Let's say you have this equation: $$x^3 - x^2 - 14x + 24 = 0$$ You have to find the roots. It's pretty difficult to do, but if we are told that 2 is a solution, we can use synthetic division to get this smaller equation $$x^2 + x - 12 = 0$$ Reduction of order does the same thing. We are told one solution to the differential equation, so we can reduce the order of the differential equation to find the other solution(s).

Method

If we know one solutions is $y_1(t)$, we plug in $v(t)y_1(t)$ for the second solution.

Example

$$y'' - \dfrac{1}{t}y' + \dfrac{1}{t^2}y = 0$$ $$y_1(t) = t \text{ is a solution. }$$ We plug in $y_2 = vt$. $$y_2 = vt$$ $$y_2' = v't + v$$ $$y_2'' = v''t + 2v'$$ We plug it back in $$v''t + 2v' - \dfrac{1}{t}\left(v't + v\right) + \dfrac{1}{t^2}vt = 0$$ $$v''t + 2v' - \dfrac{1}{t}\left(v't + v\right) + \dfrac{1}{t^2}vt = 0$$ Let's rewrite this: $$v''t + 2v' - \dfrac{1}{t}v't + \left(vt'' - \dfrac{1}{t}vt' + \dfrac{1}{t^2}vt\right) = 0$$ The part in parentheses equals 0 by our given equation. In fact, we know we can always ignore all terms that involve $v$. $$v''t + 2v' - \dfrac{1}{t}v't = 0$$ $$v''t + v' = 0$$ If we let $w = v'$, this becomes a separable differential equation. $$w = -w't \to \int \dfrac{1}{w}\, \mathrm{d}w = \int \dfrac{-1}{t}\, \mathrm{d}t$$ $$ln|w| = -ln|t| \to w = \dfrac{1}{t} \to v = ln(t)$$ So, our second solution is $vt = \boxed{tln(t)}$

Example 2

$$y''' - y'' - 14y' + 24y = 0$$ In theory, we could plug in $e^r$ and then we get $e^r\left(r^3 - r^2 - 14r + 24\right) = 0$, but that's difficult to factor. We know that $e^{2x}$ is a solution. $$y = ve^{2x}$$ $$y' = v'e^{2x} + 2ve^{2x}$$ $$y'' = v''e^{2x} + 4v'e^{2x} + 4ve^{2x}$$ $$y''' = v'''e^{2x} + 6v''e^{2x} + 12v'e^{2x} + 8ve^{2x}$$ We now remove all the terms with $v$, because those will cancel out (you can plug it in to the original equation to check). $$y = 0$$ $$y' = v'e^{2x}$$ $$y'' = v''e^{2x} + 4v'e^{2x}$$ $$y''' = v'''e^{2x} + 6v''e^{2x} + 12v'e^{2x}$$ We now plug everything in. $$e^{2x}\left(v''' + 6v'' + 12v' - v'' - 4v' - 14v'\right) = 0$$ $$v''' + 5v'' - 6v' = 0$$ We now let $w = v'$ $$w'' + 5w' - 6w = 0 \to r^2 + 5r - 6 = 0$$ $$r = 1, -6 \to w = c_1e^x + c_2e^{-6x}$$ $$v = c_1e^{x} + c_3e^{5x}$$ The constant doesn't matter, because it's still a constant. $$y_2 = \left(c_1e^{x} + c_3e^{-6x}\right)e^{2x}$$ $$y_2 = c_1e^{3x} + c_3e^{-4x}$$ The solutions are $y_1$ and collective $y_2$. $$c_1e^{2x} + c_2e^{3x} + c_3e^{-4x}$$
David Witten