Lulu M's question at Yahoo! Answers (Third order linear differential equation)

[Fernando Revilla]

Here is the question:

Consider the third order linear differential equation (1) ay′′′ +by′′ +cy′ +dy = 0

1) Transform Equation (1) to a system of first order equations of the form x′ = Ax, where x ∈ R^3;

2) Find the equation that determines the eigenvalues of the coefficient matrix A; and show that this equation is the characteristic equation of (1).

Here is a link to the question:

Third order linear differential equation? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Lulu M,

$1)$ The equation has order three, so $a\neq 0$. We can expresss: $$y'''=-\dfrac{d}{a}y-\dfrac{c}{a}y'-\dfrac{b}{a}y''\qquad (E)$$ Denoting $y_1=y,y_2=y',y_3=y''$ and using $(E)$: $$\left \{ \begin{matrix}\begin{aligned}
&y'_1=y'=y_2\\&y'_2=y''=y_3\\&y'_3=y'''=-\dfrac{d}{a}y_1-\dfrac{c}{a}y_2-\dfrac{b}{a}y_3
\end{aligned}\end{matrix}\right.$$ Equivalently:

$$\begin{bmatrix}y'_1\\y'_2\\y'_3\end{bmatrix}=
\begin{bmatrix}{\;\;0}&{\;\;1}&{\;\;0}\\{\;\;0}&{ \;\;0}&{\;\;1}\\{-d/a}&{-c/a}&{-b/a}\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix} \Leftrightarrow Y'=AY$$ $2)$ The equation that determines the eigenvalues of $A$ is: $$\begin{aligned}\det (A-\lambda I)&=\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;0}\\{\;\;0}&{-\lambda}&{\;\;1}\\{-\frac{d}{a}}&{-\frac{c}{a}}&{-\frac{b}{a}-\lambda}\end{vmatrix}\\&=-\lambda^3-\dfrac{b}{a}\lambda^2-\dfrac{d}{a}-\dfrac{c}{a}\lambda=0\\&\Leftrightarrow a\lambda^3+b\lambda^2+c\lambda+d=0\end{aligned}$$ Now, we can conclude.