# second order homogeneous differential equation method

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?

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.

• excuse me, but the solution is $(e^{(\alpha + i \beta)t},e^{(\alpha - i \beta)t})$
– Anne
Feb 17 '19 at 10:47
• It's a typo, my bad. I just fixed it. Feb 17 '19 at 10:47
• ok, but why the two coefficient are c1+c2 and c1-c2 and not two general reals k1 and k2 ?
– Anne
Feb 17 '19 at 10:52
• 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 $K_1 \cdot F(x) + K_2 \cdot G(x)$. Feb 17 '19 at 10:58
• 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?
– Anne
Feb 17 '19 at 12:29

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