Write an Equation given a solution to an ODE

[member 428835]

Homework Statement

Give an example of a system of differential equations for which $(t,1)$ is a solution.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I thought to initial pose the system as an eigenvalue problem $\vec{x}' = A \vec{x}$. However, $(t,1)$ is generally not an eigenvalue. Any ideas?

Would any system work you think, not necessarily $\vec{x}' = A \vec{x}$?

Thanks!

 

[Mark44]

Homework Statement

Give an example of a system of differential equations for which $(t,1)$ is a solution.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I thought to initial pose the system as an eigenvalue problem $\vec{x}' = A \vec{x}$. However, $(t,1)$ is generally not an eigenvalue. Any ideas?

Would any system work you think, not necessarily $\vec{x}' = A \vec{x}$?

What about this?
$\frac{dx}{dt} = t$
$\frac{dy}{dt} = 1$
Can you solve for x and y?

 

[LCKurtz]

Homework Statement

Give an example of a system of differential equations for which $(t,1)$ is a solution.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I thought to initial pose the system as an eigenvalue problem $\vec{x}' = A \vec{x}$. However, $(t,1)$ is generally not an eigenvalue. Any ideas?

Would any system work you think, not necessarily $\vec{x}' = A \vec{x}$?

Thanks!

Two questions:
1. Can you come up with a second order linear DE with those two solutions?
2. Do you know how to write a second order as a 2 by 2 matrix system?

 

[pasmith]

Homework Statement

Give an example of a system of differential equations for which $(t,1)$ is a solution.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I thought to initial pose the system as an eigenvalue problem $\vec{x}' = A \vec{x}$. However, $(t,1)$ is generally not an eigenvalue. Any ideas?

Would any system work you think, not necessarily $\vec{x}' = A \vec{x}$?

Thanks!

Rather than guessing, it is easiest to work backwards from the required solution until you have an appropriate system. Here you just need "a system of differential equations" (with no additional restrictions on the type of system) which admits the required solution, and differentiating the required solution once will get you "a system of differential equations" which admits the required solution.

That said, there is a system of the form $\vec{x}' = A \vec {x}$; since $1 = e^{0t}$ and $t = te^{0t}$ you are looking for a matrix which has two zero eigenvalues but is not the zero matrix.