Learn Laplace Transform Equations with Plainmath

The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.
${\displaystyle x_1^{{ }^{\prime }}=2x_1+x_2-x_3}$
${\displaystyle x_2^{{ }^{\prime }}=-4x_1-3x_2-x_3}$
${\displaystyle x_3^{{ }^{\prime }}=4x_1+4x_2+2x_3}$
How to find the eigenvalues by using the fact that ${\displaystyle |A-\lambda I|=0}$, how do it by inspection?

Imagine you have the matrix, $$A=(\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ -1& -1& 2\end{array})$$
By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

Obtaining Differential Equations from Functions
${\displaystyle \frac{dy}{dx}=x^2-1}$
is a first order ODE,
${\displaystyle \frac{d^{2}y}{{dx}^2}+2{\left(\frac{dy}{dx}\right)}^2+y=0}$
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
${\displaystyle y=e^x(A\cos x+B\sin x)}$ and the steps that I followed are as follows.
${\displaystyle \frac{dy}{dx}=e^x(A\cos x+B\sin x)+e^x(-A\sin x+B\cos x)}$
${\displaystyle =y+e^x(-A\sin x+B\cos x)}$ (1)
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)}$
${\displaystyle =\frac{dy}{dx}+(\frac{dy}{dx}-y)-y}$
using the orginal function and (1). Finally,
${\displaystyle \frac{d^{2}y}{{dx}^2}-2\frac{dy}{dx}+2y=0}$,
which is the required differential equation.
Similarly, if the function is $y=(A\cos 2t+B\sin 2t)$, the differential equation that I get is
${\displaystyle \frac{d^{2}y}{{dx}^2}+4y=0}$
following similar steps as above.