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

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

• I think you have a typo in the full solution - should your $C_1$ be outside the bracket? Jun 19 at 11:51
• Thanks corrected Jun 19 at 11:52
• See en.wikipedia.org/wiki/…: “which may be rewritten in real terms ...” Jun 19 at 12:05

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!

• It does.. Thank you! Jun 20 at 8:31