Non-Constant Coefficients: Reduction of Order

Second-order linear equations with non-constant coefficients don't always have solutions that can be expressed in ``closed form'' using the functions we are familiar with. However, if you know one nonzero solution of the homogeneous equation you can find the general solution (both of the homogeneous and non-homogeneous equations).

The basic idea is to write a solution as $y = y_1 u$ where $u$ is some function of $x$. What equation must $u$ satisfy?

To find out, we plug in $y = y_1 u$ to the left side of the differential equation.

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil... ...+ y_1 u'' \cr & = (2 y'_1 + p y_1) u' + y_1 u'' \cr\crcr}}$$

We will still have a second order linear differential equation for $u$, but since this only depends on $u'$ and $u''$ we can think of it as a first-order linear differential equation for $v = u'$. We know how to solve first-order linear differential equations (assuming we can do the required integrations), so we will be able to find $v$. Then we integrate to get $u$, and we have our solution $y = y_1 u$.

Example: Solve the non-homogeneous equation

$$y'' - \frac{1}{x} y' + \frac{1-x}{x} y =4 x {\rm e}^x$$

for $x > 0$, given that $$y_1 = {\rm e}^x$$ is one solution of the homogeneous equation $$y'' - \frac{1}{x} y' + \frac{1-x}{x} y = 0$$.

Taking $y = {\rm e}^x u$, we have

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil... ...2 - \frac{1}{x}\right) u'\right) = 4 x {\rm e}^x \cr\crcr}}$$

Thus the equation for $v = u'$ is

$$v' + \left(2-\frac{1}{x}\right) v = 4 x$$

The integrating factor is

$$\mu(x) = \exp \int (2 - 1/x) dx = \exp(2 x - \ln x) = {\rm e}^{2x}/x$$

so the general solution is

$$v = 4 x {\rm e}^{-2x} \left(\int {\rm e}^{2x} dx + C_1\right) = 2 x + 4 C_1 x {\rm e}^{-2x}$$

Integrating, we get

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil... ...}^x + C_1 (-1 - 2 x) {\rm e}^{-x} + C_2 {\rm e}^x \cr\crcr}}$$

Note that we get the general solution: the second fundamental solution $$y_2 = (-1-2x) {\rm e}^{-x}$$ of the homogeneous equation comes from the constant $C_1$ in the integration that gives us $v$. The constant $C_2$ in the last integration leads to the first fundamental solution (which we already knew).

Here's an example with a constant-coefficient equation. The idea is the same, but the Exponential Shift technique makes it easier. Of course, there is still no guarantee that we will be able to find the integrals in closed form.

Example: Find a solution of the differential equation

$$y'' - 4 y = \frac{{\rm e}^{-2t}}{1+{\rm e}^{-4t}}$$

The roots of the characteristic polynomial are $\pm 2$. So let's look for a particular solution of the form $$y = {\rm e}^{2t} u$$. Exponential Shift gives us

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hf... ... &= (D^2 + 4 D) u = \frac{e^{-4t}}{1+{\rm e}^{-4t}}\cr\crcr}}$$

Writing $u' = v$, we have $$v' + 4 v = {\rm e}^{-4t}/(1+{\rm e}^{-4t})$$. The integrating factor is $\mu = \exp(4 t)$. The solution to this first-order linear equation is

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil... ...-4t) \left(\frac{\ln({\rm e}^{4t}+1)}{4}+C\right)\cr \crcr}}$$

We just need a particular solution, so take $C = 0$. Integrating (it's not an easy integral):

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hf... ...{ 1 + {\rm e}^{-4t}}{16} \ln({\rm e}^{4t}+1) + C'\cr \crcr}}$$

Again, we can ignore the constant $C'$, and our particular solution is

$$y = \frac{t}{4} {\rm e}^{2t} - \frac{ {\rm e}^{2t} + {\rm e}^{-2t}}{16} \ln({\rm e}^{4t}+1)$$