I was reviewing some differential equations' topics, specifically, homogeneous linear equations with constant coefficients, from Zill & Cullen's textbook in differential equations (chapter 4, section 3)

In such section of the textbook, I came across the case for conjugate complex roots in the auxiliary equation (to solve DE's with constant coefficients), which states that the general solution (a linear combination of $y_1$ and $y_2$) for a DE with constant coefficients, whose auxiliary equation has conjugate complex roots is the following: \begin{align} y=c_1e^{\alpha x} \cos \beta x +c_2e^{\alpha x} \sin \beta x = e^{\alpha x} \left( c_1 \cos \beta x + c_2 \sin \beta x \right) \end{align}

Which uses the Euler's formula for sines and cosines, however, a bit earlier in the text (the paragraph before the general solution), it is stated that $y_1$ and $y_2$ are the following:

\begin{align} y_1&=e^{\alpha x} \left( e^{i\beta x} + e^{-i\beta x} \right)= 2e^{\alpha x} \cos \beta x \\ y_2&=e^{\alpha x} \left( e^{i\beta x} - e^{-i\beta x} \right)= 2ie^{\alpha x} \sin \beta x \end{align}

From $y_1$ and $y_2$ it is visible the presence of a 2 with $y_1$ and $2i$ with $y_2$, but, none of those are present in the general solution $y$.

The question:

How is this possible? Considering that a solution in the real domain is to be found, and there are complex elements in the individual solutions.

Are there any other considerations left out?

  • $\begingroup$ The general solution has been written as $c_1\frac12y_1+c_2\frac{1}{2i}y_2$ so the real roots are those with $c_1,\,c_2\in\Bbb R$. $\endgroup$
    – J.G.
    Commented Mar 2, 2023 at 17:11
  • $\begingroup$ @AnneBauval I meant that the general solution is a linear combination of $y_1$ and $y_2$, thanks for the observation! $\endgroup$
    – Carlos R.
    Commented Mar 2, 2023 at 17:19

1 Answer 1


$(y_1,y_2)$ is a fundamental system of complex solutions (i.e. a basis of the $\Bbb C$-vector space of $\Bbb C$-valued solutions) and $y_1/2,y_2/(2i)$ are real, hence $(y_1/2,y_2/(2i))$ is a fundamental system of real solutions.

  • $\begingroup$ So, considering the general solution is a linear combination, with constants $c_1$ and $c_2$, does $c_1$ and $c_2$ have the form $c_1= \frac{a}{2}$ and $c_2= \frac{b}{2i}$, respectively? $\endgroup$
    – Carlos R.
    Commented Mar 2, 2023 at 17:42
  • $\begingroup$ $a,b$ were mentionned nowhere. I'd rather simply say that if $(u,v)$ is a fundamental system of $\Bbb K$-valued solutions then the general $\Bbb K$-valued solution is $c_1u+c_2v$ with $c_1,c_2\in\Bbb K,$ and apply this either to $(u,v)=(y_1,y_2)$ and $\Bbb K=\Bbb C$, or to $\Bbb K=\Bbb R$ (or $\Bbb C$) and $(u,v)=(y_1/2,y_2/(2i)).$ $\endgroup$ Commented Mar 2, 2023 at 17:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .