How to get the (real function) solution of a differential equation when the characteristic roots are complex numbers

Question

When solving a differential formula, say, homogeneous, order two, with constant coefficients, we look at the solutions of the characteristic polynomial, say we got $r_1 \neq r_2$.

We say that the solution is: $y(x) = C_1 e^{r_1 x}+C_2 e^{r_2 x}$

So suppose the roots are complex, say, $r_1=a+bi,r_2=a-bi$.

Then the solution should be: $y(x) = e^{a x} [C_1 ( cos (bx) + i sin (bx))+C_2 (cos (bx) - i sin (bx))]$

However, this solution is not a real function.

So, what is the solution in this case of two complex roots?

Thank you

Answer 1

It looks from your expression as though you only have one unknown constant; since the solution is real, shouldn't we just have $C_1=C_2$ so the imaginary terms cancel?

What you might be missing is the fact that $C_1$ and $C_2$ can also be complex. Say $C_n = u_n + iv_n$ for $n=1,2$; then $$y(x)=e^{ax} \left[\left(u_1+iv_1\right)\left(\cos bx +i\sin bx \right)+\left(u_2+iv_2\right)\left(\cos bx -i\sin bx \right)\right]$$

We do want imaginary parts to vanish for a real solution; so $$u_1 \sin bx + v_1 \cos bx - u_2 \sin bx + v_2 \cos bx = 0$$

for all $x$. When $bx=0$, this is $$v_1 + v_2 = 0$$

When $bx=\frac{\pi}{2}$, we get $$u_1 - u_2 = 0$$

ie $u_2=u_1$ and $v_2=-v_1$. Dropping the subscripts,

$$y(x)=e^{ax} \left[\left(u+iv\right)\left(\cos bx +i\sin bx \right)+\left(u-iv\right)\left(\cos bx -i\sin bx \right)\right]$$

ie

$$y(x)=2e^{ax} \left[u\cos bx -v\sin bx\right]$$

You can either leave the result in this form (with $u$ and $v$ as your constants) or rewrite as

$$y(x)=Re^{ax} \cos (bx+\theta)$$

(and you can choose which depending on the situation). Hope that helps!