Characteristic equation with negative discriminant We consider the homogeneous second-order linear differential equation with constant coefficients, written in the form \[a\,\frac{\dd^2y}{\dd t^2}+b\,\frac{\dd y}{\dd t}+c\,y(t)=0\] The corresponding characteristic equation is \[a\,\lambda^2+b\,\lambda+c=0\]

We restrict ourselves to the case where the discriminant \(D=b^2-4ac\) is negative. Then there are, according to the quadratic formula, two complex roots, say \[\lambda_1=\frac{-b+\mathrm{i}\,\sqrt{-D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\mathrm{i}\,\sqrt{-D}}{2a}\] The following two functions are solutions of the differential equation \[y_1(t)=e^{\lambda_1t}\qquad\text{and}\qquad y_2(t)=e^{\lambda_2t}\] In fact, any solution can be written as a linear combination of these two "basic complex solutions".

You might feel more at ease when we do not work with complex solutions, but only with real numbers and real functions. This can be arranged by working with special combinations of the complex solutions. Suppose that the standard form of the complex number \(\lambda_1\) is equal to \(\alpha+\mathrm{i}\,\beta\), then the complex number \(\lambda_2\) is equal to \(\alpha-\mathrm{i}\,\beta\) and \[\alpha=-\frac{b}{2a}\qquad\text{and}\qquad \beta=\frac{\sqrt{4ac-b^2}}{2a}\] The complex numbers \(\lambda_1\) and \(\lambda_2\) are conjugates of each other. Then: \[\begin{aligned}\frac{e^{\lambda_1t} +e^{\lambda_2t}}{2} &= \frac{e^{(\alpha+\mathrm{i}\,\beta)t} +e^{(\alpha-\mathrm{i}\,\beta)t}}{2}\\ \\ &= \e^{\alpha t}\,\frac{e^{\mathrm{i}\,\beta\, t} +e^{-\mathrm{i}\,\beta\, t}}{2}\\ \\ &= e^{\alpha t}\cos(\beta t)\end{aligned}\] Here you have the first real solution to the differential equation. Another combination gives a second real solution \[\begin{aligned}\frac{e^{\lambda_1t} -e^{\lambda_2t}}{2\mathrm{i}} &= \frac{e^{(\alpha+\mathrm{i}\,\beta)t} -e^{(\alpha-\mathrm{i}\,\beta)t}}{2\mathrm{i}}\\ \\ &= \e^{\alpha t}\,\frac{e^{\mathrm{i}\,\beta\, t} +e^{-\mathrm{i}\,\beta\, t}}{2\mathrm{i}}\\ \\ &= e^{\alpha t}\sin(\beta t)\end{aligned}\] A linear combination of these two real solutions gives the general solution \[y(t)=e^{\alpha t}\cdot\bigl(A\cos(\beta t)+B\sin(\beta t)\bigr)\] with constants \(A\) and \(B\). These constants are determined by boundary conditions.