How Can I Solve a Linear System of Differential Equations Without Determinants?

In summary, the conversation discussed solving a linear system of differential equations using the concept of eigenvectors without determinants. The solution involved finding the eigenvalues and eigenvectors, but there were some mistakes in the eigenvector calculations. The conversation also mentioned a bug in the forum software regarding incorrect images appearing for rendered LaTeX.
  • #1
mrxtothaz
14
0
I am solving for a linear system of differential equations. Since LaTeX never works out for me on this forum, I have written out my problem and my solution on LaTeX and made an image of it: http://i.imgur.com/ncra.jpg .

I basically have to solve the system of equations and express it as a diagonal matrix. For this, I use the concept of eigenvectors (this is done without determinants). I ran into some difficulties and I am not confident in my solution, so I would really appreciate any feedback on my solution.

Thanks in advance.
 
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  • #2
There's a bug in the forum software where incorrect images sometimes appear for the rendered LaTeX in previews. The work-around is to reload the page. The correct images will show up then.

Your problem is to solve the system

[tex]\begin{pmatrix}\dot{x}_1 \\ \dot{x}_2\end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}[/tex]

You found the eigenvalues are [itex]\lambda = -1[/itex] and [itex]\lambda=3[/itex]. So far, so good.

Then you wrote the eigenvector for [itex]\lambda = -1[/itex] is

[tex]\vec{v}_1 = t\begin{pmatrix}1 \\ 1\end{pmatrix}[/tex]

The factor of t shouldn't be there.

Next you said the eigenvectors for [itex]\lambda = 3[/itex] are

[tex]\vec{v}_2 = t\begin{pmatrix}-1 \\ 1\end{pmatrix} \hspace{0.5in} \vec{v}_3 = t\begin{pmatrix}1 \\ -1\end{pmatrix}[/tex]

Again, the factor of t shouldn't be there. Also, note that the first vector is just a multiple of the second vector, so they're essentially the same eigenvector.
 

What is a linear system?

A linear system is a set of equations that can be represented graphically as a system of lines. Each equation in the system contains variables that must be solved for in order to find a solution that satisfies all of the equations.

What is the process for solving a linear system?

The process for solving a linear system involves transforming the equations into a simpler form, such as standard form or slope-intercept form, and then using algebraic techniques, such as substitution or elimination, to solve for the unknown variables.

What are the different methods for solving a linear system?

There are three main methods for solving a linear system: graphing, substitution, and elimination. Graphing involves plotting the equations on a coordinate plane and finding the point of intersection. Substitution involves solving one equation for a variable and then substituting the expression into the other equation. Elimination involves adding or subtracting equations to eliminate one of the variables.

When is a linear system considered consistent or inconsistent?

A linear system is considered consistent when it has at least one solution that satisfies all of the equations. It is considered inconsistent when there is no solution that satisfies all of the equations. This can happen when the equations are parallel lines or when they do not intersect on the coordinate plane.

How can solving a linear system be applied in real life?

Solving a linear system can be applied in many real-life situations, such as calculating the cost and revenue for a business, determining the optimal mix of ingredients in a recipe, or finding the intersection point of two moving objects. It is a fundamental tool in fields such as physics, engineering, economics, and statistics.

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