# Complex solutions of ordinary differential equation of order 2

I'm trying to understand the math notes detailing the way to obtain the complex solution:

If the problem is homogeneous we have :

$$ay''(t) + by'(t) + cy(t) = 0$$

to know all solutions for this use the ansatz

$$y(t) = e^{\lambda t}$$

and plug it into the ODE to find

$$ay''(t) + by'(t) + cy(t) = a \lambda^2e^{\lambda t} + b\lambda e^{\lambda t} + ce^{\lambda t} = 0$$

This problem can be solved using the characteristic equation:

$$a\lambda^2 + b\lambda +c = 0$$

Dependeing on the value of the discriminant $$D = b^2 - 4ac$$ we find different types of solutions:

$$1.$$ $$D < 0$$: $$\lambda_1 \neq \lambda_2$$ and both real.
Then we have two linearly independent solutions

$$y_1(t) = e^{\lambda_1t} \quad \text{and}\quad y_2(t) = e^{\lambda_2t}$$ and thus the general solution is

$$y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{\lambda_1t} + c_2e^{\lambda_2t}$$

$$2.$$ $$D = 0$$: $$\lambda_1 = \lambda_2 \in \mathbb{R}$$
One solution is immediate from the previous case

$$y_1(t) = e^{\lambda_1t}$$

One can verify that the second solution is given by

$$y_2(t) = te^{\lambda_1t}$$ Thus the general solution is

$$y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{\lambda_1t} (c_1 +tc_2)$$

$$3.$$ $$D < 0$$: $$\lambda_{1,2} = \alpha \pm i\beta$$ with $$\alpha$$, $$\beta \in \mathbb{R}$$ and $$\beta \neq 0$$
Using the idea from the first case we end up with complex sotluions

$$u_1(t) =e^{\lambda_1t} = e^{(\alpha + i \beta)t} = e^{\alpha t}(\cos(\beta t) + i \sin(\beta t)) \tag{1}\label{eq1}$$ $$u_2(t) =e^{\lambda_2t} = e^{(\alpha + i \beta)t} = e^{\alpha t}(\cos(\beta t) - i \sin(\beta t)) \tag{2}\label{eq2}$$

My Question:

I guess the statement $$(1)$$ and $$(2)$$ have to do with the properties of complex numbers. However can you give me a link to understand those $$2$$ statements.

• That's Euler's formula: en.wikipedia.org/wiki/Euler%27s_formula – Ethan Bolker Oct 27 '18 at 15:07
• Actually, it's 1 and 3 which are easily linked through complex numbers: $y(t)=Ae^{\lambda_1t}+Be^{\lambda_2 t}$ works for both of them, and there is no real reason to consider them two distinct cases. As for how to link (2) to the other two, I asked about that here. You can see what you think about the answers given there. – Arthur Oct 27 '18 at 15:07
• thank you @EthanBolker! – ecjb Oct 27 '18 at 15:08

$$e^{i\theta}=\cos\theta+i\sin\theta$$
However, to summarize, the function $$f(t)=e^{i\beta t}$$ for some constant $$\beta$$ is a complex function that starts at $$f(0)=1$$ and then rotates around the unit circle at $$\beta$$ radians per second (assuming $$t$$ is a variable of time in seconds). Then, the function $$f(t)=e^{(\alpha+i\beta)t}=e^{\alpha t}e^{i\beta t}$$ for $$\alpha > 0$$ is a function that starts at $$f(0)=1$$ and then rotates around the unit circle at $$\beta$$ radians per second while also getting exponentially farther away from the origin because of the exponentially growing magnitude $$e^{\alpha t}$$. However, if $$\alpha < 0$$, then $$f(t)=e^{\alpha t}e^{i\beta t}$$ gets closer and closer to the origin because of the exponentially decaying magnitude $$e^{\alpha t}$$.