Differential Equations - Second Order: Complex Roots

In this section we will be looking at solutions to the differential equation

$$a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$$

in which roots of the characteristic equation,

$$a{r}^2+br+c=0$$

are complex roots in the form $r_{1,2}=\lambda \pm \mu i$.

Now, recall that we arrived at the characteristic equation by assuming that all solutions to the differential equation will be of the form

$$y\left(t\right)=e^{rt}$$

Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

$$y_1\left(t\right)=e^{\left(\lambda +\mu i\right)t}\text{and}{y}_2\left(t\right)=e^{\left(\lambda -\mu i\right)t}$$

Now, these two functions are “nice enough” (there’s those words again… we’ll get around to defining them eventually) to form the general solution. We do have a problem however. Since we started with only real numbers in our differential equation we would like our solution to only involve real numbers. The two solutions above are complex and so we would like to get our hands on a couple of solutions (“nice enough” of course…) that are real.

To do this we’ll need Euler’s Formula.

$$e^{i\theta }=\cos \theta +i\sin \theta$$

A nice variant of Euler’s Formula that we’ll need is.

$$e^{-i\theta }=\cos \left(-\theta \right)+i\sin \left(-\theta \right)=\cos \theta -i\sin \theta$$

Now, split up our two solutions into exponentials that only have real exponents and exponentials that only have imaginary exponents. Then use Euler’s formula, or its variant, to rewrite the second exponential.

$$\begin{array}{cc}{y}_1\left(t\right)& =e^{\lambda t}{e}^{i\mu t}=e^{\lambda t}\left(\cos \left(\mu t\right)+i\sin \left(\mu t\right)\right)\\ y_2\left(t\right)& =e^{\lambda t}{e}^{-i\mu t}=e^{\lambda t}\left(\cos \left(\mu t\right)-i\sin \left(\mu t\right)\right)\end{array}$$

This doesn’t eliminate the complex nature of the solutions, but it does put the two solutions into a form that we can eliminate the complex parts.

Recall from the basics section that if two solutions are “nice enough” then any solution can be written as a combination of the two solutions. In other words,

$$y\left(t\right)=c_1{y}_1\left(t\right)+c_2{y}_2\left(t\right)$$

will also be a solution.

Using this let’s notice that if we add the two solutions together we will arrive at.

$$y_1\left(t\right)+y_2\left(t\right)=2e^{\lambda t}\cos \left(\mu t\right)$$

This is a real solution and just to eliminate the extraneous 2 let’s divide everything by a 2. This gives the first real solution that we’re after.

$$u\left(t\right)=\frac{1}{2}{y}_1\left(t\right)+\frac{1}{2}{y}_2\left(t\right)=e^{\lambda t}\cos \left(\mu t\right)$$

Note that this is just equivalent to taking

$$c_1=c_2=\frac{1}{2}$$

Now, we can arrive at a second solution in a similar manner. This time let’s subtract the two original solutions to arrive at.

$$y_1\left(t\right)-y_2\left(t\right)=2i{e}^{\lambda t}\sin \left(\mu t\right)$$

On the surface this doesn’t appear to fix the problem as the solution is still complex. However, upon learning that the two constants, $c_1$ and $c_2$ can be complex numbers we can arrive at a real solution by dividing this by $2i$. This is equivalent to taking

$$c_1=\frac{1}{2i}\text{and}{c}_2=-\frac{1}{2i}$$

Our second solution will then be

$$v\left(t\right)=\frac{1}{2i}{y}_1\left(t\right)-\frac{1}{2i}{y}_2\left(t\right)=e^{\lambda t}\sin \left(\mu t\right)$$

We now have two solutions (we’ll leave it to you to check that they are in fact solutions) to the differential equation.

$$u\left(t\right)=e^{\lambda t}\cos \left(\mu t\right)\text{and}v\left(t\right)=e^{\lambda t}\sin \left(\mu t\right)$$

It also turns out that these two solutions are “nice enough” to form a general solution.

So, if the roots of the characteristic equation happen to be $r_{1,2}=\lambda \pm \mu i$ the general solution to the differential equation is.

$$y\left(t\right)=c_1{e}^{\lambda t}\cos \left(\mu t\right)+c_2{e}^{\lambda t}\sin \left(\mu t\right)$$

Let’s take a look at a couple of examples now.

Example 1 Solve the following IVP.$$y^{\prime \prime }-4y^{\prime }+9y=0y\left(0\right)=0y^{\prime }\left(0\right)=-8$$

The characteristic equation for this differential equation is.
$$r^2-4r+9=0$$The roots of this equation are $r_{1,2}=2\pm \sqrt{5}i$. The general solution to the differential equation is then.
$$y\left(t\right)=c_1{e}^{2t}\cos \left(\sqrt{5}t\right)+c_2{e}^{2t}\sin \left(\sqrt{5}t\right)$$Now, you’ll note that we didn’t differentiate this right away as we did in the last section. The reason for this is simple. While the differentiation is not terribly difficult, it can get a little messy. So, first looking at the initial conditions we can see from the first one that if we just applied it we would get the following.
$$0=y\left(0\right)=c_1$$In other words, the first term will drop out in order to meet the first condition. This makes the solution, along with its derivative
$$\begin{array}{cc}y\left(t\right)& =c_2{e}^{2t}\sin \left(\sqrt{5}t\right)\\ y^{\prime }\left(t\right)& =2c_2{e}^{2t}\sin \left(\sqrt{5}t\right)+\sqrt{5}{c}_2{e}^{2t}\cos \left(\sqrt{5}t\right)\end{array}$$A much nicer derivative than if we’d done the original solution. Now, apply the second initial condition to the derivative to get.
$$-8=y^{\prime }\left(0\right)=\sqrt{5}{c}_2\Rightarrow c_2=-\frac{8}{\sqrt{5}}$$The actual solution is then.
$$y\left(t\right)=-\frac{8}{\sqrt{5}}{e}^{2t}\sin \left(\sqrt{5}t\right)$$

Example 2 Solve the following IVP.$$y^{\prime \prime }-8y^{\prime }+17y=0y\left(0\right)=-4y^{\prime }\left(0\right)=-1$$

The characteristic equation this time is.
$$r^2-8r+17=0$$The roots of this are $r_{1,2}=4\pm i$. The general solution as well as its derivative is
$$\begin{array}{cc}y\left(t\right)& =c_1{e}^{4t}\cos \left(t\right)+c_2{e}^{4t}\sin \left(t\right)\\ y^{\prime }\left(t\right)& =4c_1{e}^{4t}\cos \left(t\right)-c_1{e}^{4t}\sin \left(t\right)+4c_2{e}^{4t}\sin \left(t\right)+c_2{e}^{4t}\cos \left(t\right)\end{array}$$Notice that this time we will need the derivative from the start as we won’t be having one of the terms drop out. Applying the initial conditions gives the following system.
$$\begin{array}{cc}-4& =y\left(0\right)=c_1\\ -1& =y^{\prime }\left(0\right)=4c_1+c_2\end{array}$$Solving this system gives $c_1=-4$ and $c_2=15$. The actual solution to the IVP is then.
$$y\left(t\right)=-4e^{4t}\cos \left(t\right)+15e^{4t}\sin \left(t\right)$$

Example 3 Solve the following IVP.$$4y^{\prime \prime }+24y^{\prime }+37y=0y\left(\pi \right)=1y^{\prime }\left(\pi \right)=0$$

The characteristic equation this time is.
$$4r^2+24r+37=0$$The roots of this are $r_{1,2}=-3\pm \frac{1}{2}i$. The general solution as well as its derivative is
$$\begin{array}{cc}y\left(t\right)& =c_1{e}^{-3t}\cos \left(\frac{t}{2}\right)+c_2{e}^{-3t}\sin \left(\frac{t}{2}\right)\\ y^{\prime }\left(t\right)& =-3c_1{e}^{-3t}\cos \left(\frac{t}{2}\right)-\frac{c_1}{2}{e}^{-3t}\sin \left(\frac{t}{2}\right)-3c_2{e}^{-3t}\sin \left(\frac{t}{2}\right)+\frac{c_2}{2}{e}^{-3t}\cos \left(\frac{t}{2}\right)\end{array}$$Applying the initial conditions gives the following system.
$$\begin{array}{cc}1& =y\left(\pi \right)=c_1{e}^{-3\pi }\cos \left(\frac{\pi }{2}\right)+c_2{e}^{-3\pi }\sin \left(\frac{\pi }{2}\right)=c_2{e}^{-3\pi }\\ 0& =y^{\prime }\left(\pi \right)=-\frac{c_1}{2}{e}^{-3\pi }-3c_2{e}^{-3\pi }\end{array}$$Do not forget to plug the $t=\pi$ into the exponential! This is one of the more common mistakes that students make on these problems. Also, make sure that you evaluate the trig functions as much as possible in these cases. It will only make your life simpler. Solving this system gives
$$c_1=-6e^{3\pi }{c}_2=e^{3\pi }$$The actual solution to the IVP is then.
$$\begin{array}{cc}y\left(t\right)& =-6e^{3\pi }{e}^{-3t}\cos \left(\frac{t}{2}\right)+e^{3\pi }{e}^{-3t}\sin \left(\frac{t}{2}\right)\\ y\left(t\right)& =-6e^{-3\left(t-\pi \right)}\cos \left(\frac{t}{2}\right)+e^{-3\left(t-\pi \right)}\sin \left(\frac{t}{2}\right)\end{array}$$

Let’s do one final example before moving on to the next topic.

Example 4 Solve the following IVP.$$y^{\prime \prime }+16y=0y\left(\frac{\pi }{2}\right)=-10y^{\prime }\left(\frac{\pi }{2}\right)=3$$

The characteristic equation for this differential equation and its roots are.
$$r^2+16=0\Rightarrow r=\pm 4i$$Be careful with this characteristic polynomial. One of the biggest mistakes students make here is to write it as,
$$r^2+16r=0$$
The problem is that the second term will only have an $r$ if the second term in the differential equation has a $y^{\prime }$ in it and this one clearly does not. Students however, tend to just start at $r^2$ and write times down until they run out of terms in the differential equation. That can, and often does mean, they write down the wrong characteristic polynomial so be careful.
Okay, back to the problem.
The general solution to this differential equation and its derivative is.
$$\begin{array}{cc}y\left(t\right)& =c_1\cos \left(4t\right)+c_2\sin \left(4t\right)\\ y^{\prime }\left(t\right)& =-4c_1\sin \left(4t\right)+4c_2\cos \left(4t\right)\end{array}$$Plugging in the initial conditions gives the following system.
$$\begin{array}{cc}-10& =y\left(\frac{\pi }{2}\right)=c_1{c}_1=-10\\ 3& =y^{\prime }\left(\frac{\pi }{2}\right)=4c_2{c}_2=\frac{3}{4}\end{array}$$So, the constants drop right out with this system and the actual solution is.
$$y\left(t\right)=-10\cos \left(4t\right)+\frac{3}{4}\sin \left(4t\right)$$