Constant Coefficients: Complex Roots

We are now going to apply our knowledge of complex numbers to solving second order constant coefficient linear differential equations. The main principles are the same as what we considered previously, except that we now allow complex roots (and know how to interpret $\exp(r x)$ when $r$ is complex).

Example: Solve $(D^2 + 2 D + 5) y = 0$.

The characteristic equation $r^2 + 2 r + 5 = 0$ has roots $r = -1 \pm 2 i$ so we have a fundamental set of solutions: $\exp((-1+2i) x) = \exp(-x)(\cos 2x + i \sin 2x)$ and $\exp((-1-2i) x) = \exp(-x)(\cos 2x - i \sin 2x)$. Note that the two complex roots are complex conjugates of each other, and also the corresponding solutions of the differential equation are complex conjugates of each other.

Now usually we expect our solutions to differential equations to be real. Certainly if $y$ is supposed to represent some real physical quantity, we would be interested in a real solution. Real solutions can be obtained as suitable linear combinations of our fundamental set of complex solutions. For example,

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil$... ...-1+2i) x) - i \exp((-1-2i)x) &= -2 \exp(-x) \sin 2x\cr\crcr}}$$

But a rather more convenient way to obtain real solutions is as real and imaginary parts of a complex solution:

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil$... ...cr {\rm Im }\exp((-1+2i) x) &= \exp(-x) \sin 2x \cr\crcr}}$$

What's involved here is sometimes called the ``Principle of Equating Real Parts.'' A linear operator $T$ is said to be real if $Tu$ is real whenever $u$ is real. For example, differential operators $T = D^2 + p D + q$ are real if the coefficients $p$ and $q$ are real. Then the principle says:

Let $z = u + i v$ be a complex solution of $Tz= f$, where $T$, $u$ and $v$ are real. Then $Tu = {\rm Re }f$ and $Tv = {\rm Im }f$.

Proof: $f = Tz = Tu + i Tv$, where $Tu$ and $Tv$ are real because $u$ and $v$ are real. So $Tu = {\rm Re }f$ and $Tv = {\rm Im }f$.

In particular, this says that if $r$ is a complex root of the real polynomial $r^2 + p r + q$ then ${\rm Re }\exp(r x)$ and ${\rm Im }\exp(r x)$ are solutions of the homogeneous equation $(D^2 + p D + q) y = 0$. If $r = a + b i$, these solutions are $$\exp(a x) \cos (bx)$$ and $$\exp(a x) \sin (bx)$$.

There is no need, however, to restrict ourselves to homogeneous equations.

Example: Find a particular solution of $(D^2 + D + 5) y = \exp(-x) \cos x$.

The right side is the real part of $\exp(-x) \exp(ix) = \exp((-1+i)x)$. So we can solve $(D^2 + D + 5) z = \exp((-1+i)x)$, and then take $y = {\rm Re }z$. Now for $r= -1+i$ we have $r^2 + r + 5 = 4 - i$, so

$$\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil$... ...\frac{4}{17} \cos x - \frac{1}{17} \sin x \right)\cr\crcr}}$$

At the same time we also obtain a solution of $(D^2 + D + 5) y = \exp(-x) \sin x$:

$${\rm Im }z = \exp(-x) \left( \frac{4}{17} \sin x + \frac{1}{17}\cos x\right)$$

We can also use this principle together with the Exponential Shift Theorem.

Example: Find a particular solution of $(D^2 + 2 D + 5) y = \exp(-x) \sin (2 x)$.

Let $r = -1+2i$. We want to find a particular solution of $(D^2 + 2 D + 5) z = \exp(r x)$ and then take $y = {\rm Im }z$. Since $r^2 + 2 r + 5 = 0$, we must take $z = \exp(r x) u$. According to the Exponential Shift Theorem,

$$\,\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil$... ...2) D + r^2 + 2 r + 5) u = \exp(r x)(D^2 + 4 i D) u\cr\crcr}}\,$$

so we want $(D + 4 i) D u = 1$. We can try $v = D u =$constant, and then $v = 1/(4 i) = -i/4$, so $u = -i x/4$, $z = -i \exp(r x) x/4$, and

$$y = {\rm Im }z = - \frac{x}{4} \exp(-x) \cos(2 x)$$