Quick Question: Matrix added to Differential

[rey242]

Hey guys,
I have a quick question about adding a vector to a system of differential equations.

Like U'(t)= Ax+V

Say A is a 2x2 matrix and V is a 2x1 vector.

Can you all explain how I could handle that V vector? Should somehow include it into A?

 

[Mark44]

Hey guys,
I have a quick question about adding a vector to a system of differential equations.

Like U'(t)= Ax+V

Say A is a 2x2 matrix and V is a 2x1 vector.

Can you all explain how I could handle that V vector? Should somehow include it into A?

You're not adding a vector to a system of differential equations. Ax is a 2x1 vector, and V is a 2x1 vector, so Ax + V is a 2x1 vector, as is U', which is probably a function of x, not t.

 

[lanedance]

As mentioned you haven't explained the problem very well... but guessing, do you mean
\textbf{x}(t) = (x_1(t), x_2(t))
\textbf{x}'(t) = (x_1'(t), x_2'(t))

and the differential equation is
\textbf{x}'(t) = A \textbf{x}(t) +\textbf{v}...?

if so, and if A is in invertible, try writing it as
\textbf{x}'(t) = A \textbf{x}(t) +A(A^{-1}\textbf{v})
and consider a variable change

 

[rey242]

I wrote the question a little wrong...
I meant
U'(t)=A u(t)+V
Where A is just a regular 2X2 matrix and V is a 2x1 vector.
like this

U'(t)=(1,2;3,4)U(t)+(1;e^t)

I already know how to solve systems such as
U'(t)=Au(t) by using the eigenvalues & eigenvectors of A but I haven't learned what to do in the case which you add a V into the equation

 

[Mark44]

U'(t) = AU(t) is a homogeneous system.
U'(t) = AU(t) + V is a nonhomogeneous system. What you choose for a particular solution depends a lot on what the solution is for the homogeneous problem. I don't have my DE resources handy right now, so I can't provide any more details these.

 

[rey242]

so what if it included a initial solution, would that change anything?

 

[Mark44]

Do you mean initial conditions?

 

[rey242]

Such as a vector U(0)=(1;1)

 

[Mark44]

That would enable you to find the constants associated with your homogeneous solution, that's all. IOW, having initial conditions wouldn't really change what I said in post #5.