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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.

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    $\begingroup$ That's Euler's formula: en.wikipedia.org/wiki/Euler%27s_formula $\endgroup$ – Ethan Bolker Oct 27 '18 at 15:07
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    $\begingroup$ 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. $\endgroup$ – Arthur Oct 27 '18 at 15:07
  • $\begingroup$ thank you @EthanBolker! $\endgroup$ – ecjb Oct 27 '18 at 15:08
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These equations are derived using Euler's Formula.

$$e^{i\theta}=\cos\theta+i\sin\theta$$

To learn more about why this formula works, I would recommend this BetterExplained article, which gives you a more intuitive understanding of what Euler's Formula really means.

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 $e^{\alpha t}$.

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