second order homogeneous differential equation method
Sebastian Wright 
I'm trying to understand the general method to solve the second order differential equations $u''+2au'+bu=f(t)$.
The homogeneous differential equation $u''+2au'+bu=0$ can be solved looking for a solution in the form $e^{\lambda t }$ and solving the characteristic equation $\lambda^2+2a \lambda+b=0$
that will have two solutions $\lambda_{1/2}=-a \pm \sqrt{a^2-b}$.
They can be real numbers or, more generally, complex numbers $\alpha \pm i \beta$ from which the solution of the homogeneous differentiale equation will be$$e^{\alpha + i \beta}=e^{\alpha}(cos \beta t+isen \beta t)$$ or $$e^{\alpha - i \beta}=e^{\alpha}(cos \beta t-isen \beta t)$$
The general solution of the homogenous differential equation will be $$c_1*e^{\alpha}(cos \beta t+isen \beta t)+c_2*e^{\alpha}(cos \beta t-isen \beta t)=(c_1+c_2)*e^{\alpha}cos \beta t+i(c_1-c_2)*e^{\alpha}sen \beta t$$
I'm trying to understand if $e^{\alpha}cos \beta t$ or $e^{\alpha}sen \beta$. Can someone help me to understand?
$\endgroup$2 Answers
$\begingroup$When you solve a second-order linear differential homogeneous equation, you have TWO bases of functions so that the general solution is $C_1 \cdot F(x) + C_2 G(x)$. And you find $C_1$ and $C_2$ with the boundary or initial conditions.
For this differential equation, a base of the general solution is $(e^{(\alpha + i \beta)t},e^{(\alpha - i \beta)t})$. But you can also choose other bases. Using Euler's formula, you can also use sin and cos or even sinh and cosh.
So you found that $(e^{\alpha t} cos(\beta t), i\cdot e^{\alpha t} \cdot sin(\beta t))$, which is correct aswell.
$\endgroup$ 8 $\begingroup$Excuse me, but I didn't understant completely when you wrote this:
"So you found that $(e^{\alpha t} cos(\beta t), i\cdot e^{\alpha t} \cdot sin(\beta t))$, which is correct aswell." I don't understant why you put $i$ either in the base of linearly independent vectors or in the system to find the two coefficients.
In other words is the new base $(e^{\alpha t} cos(\beta t), i\cdot e^{\alpha t} \cdot sin(\beta t))$ or $(e^{\alpha t} cos(\beta t), e^{\alpha t} \cdot sin(\beta t))$ or the system is $c_1+c_2=k_1$ and $c_1-c_2=k_2$ or $c_1+c_2=k_1$ and $(c_1-c_2)*i=k_2$?
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