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

AI Thread Summary
The discussion focuses on transforming a third order linear differential equation into a system of first order equations. The equation is expressed in terms of a matrix A, where the derivatives of y are represented as a vector. The characteristic equation for the matrix A is derived, leading to the determination of its eigenvalues. The transformation and eigenvalue equation are crucial for understanding the behavior of the original differential equation. This process highlights the relationship between the original equation and its characteristic polynomial.
Fernando Revilla
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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.
 
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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.
 
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