1
$\begingroup$

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 2

2
$\begingroup$

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 the solution is $(e^{(\alpha + i \beta)t},e^{(\alpha - i \beta)t})$ $\endgroup$
    – Anne
    Feb 17, 2019 at 10:47
  • $\begingroup$ It's a typo, my bad. I just fixed it. $\endgroup$ Feb 17, 2019 at 10:47
  • $\begingroup$ ok, but why the two coefficient are c1+c2 and c1-c2 and not two general reals k1 and k2 ? $\endgroup$
    – Anne
    Feb 17, 2019 at 10:52
  • $\begingroup$ The two coefficients $K_1$ and $K_2$ are part of the complex world. It just happens that you usually deal with the real equations first, then move to the complex world. You can just set $K_1 = c_1 + c_2$ over vice versa $ 2 c_1 = K_1 - K_2$. Another way to write a space of solution is $(F(x),G(x))$ standing for $K_1 \cdot F(x) + K_2 \cdot G(x)$. $\endgroup$ Feb 17, 2019 at 10:58
  • $\begingroup$ so it doesn't matter if the two values of the constants are linked one to the other, the important thing is that they are two that I can choose as I want, isn't it? $\endgroup$
    – Anne
    Feb 17, 2019 at 12:29
0
$\begingroup$

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 $(c_1-c_2)*i=k_2$?

$\endgroup$

You must log in to answer this question.

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