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
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ excuse me, but the solution is $(e^{(\alpha + i \beta)t},e^{(\alpha - i \beta)t})$ $\endgroup$ – Anne Feb 17 at 10:47
  • $\begingroup$ It's a typo, my bad. I just fixed it. $\endgroup$ – PackSciences Feb 17 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 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 $K_1 \cdot F(x) + K_2 \cdot G(x)$. $\endgroup$ – PackSciences Feb 17 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 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 $(c_1-c_2)*i=k_2$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.