Using Eigenvalues and Eigenvectors to solve Differential Equations

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Homework Statement



x1(t) and x2(t) are functions of t which are solutions of the system of differential equations

x(dot)1 = 4x1 + 3x2
x(dot)2 = -6x1 - 5x2

Express x1(t) and x2(t) in terms of the exponential function, given that x1(0) = 1 and x2(0) = 0

The Attempt at a Solution



I've already put this into matrix form,

4...3
-6 -5

and then

(4-λ) 3
-6 (-5-λ)

To find that the eigenvalues are -2 and 1, and then that the eigenvectors are

(a
-2a)

and

(b
-b)

but after that I'm completely stuck. I know you get into constants and such but I just can't remember how to continue.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



x1(t) and x2(t) are functions of t which are solutions of the system of differential equations

x(dot)1 = 4x1 + 3x2
x(dot)2 = -6x1 - 5x2

Express x1(t) and x2(t) in terms of the exponential function, given that x1(0) = 1 and x2(0) = 0

The Attempt at a Solution



I've already put this into matrix form,

4...3
-6 -5

and then

(4-λ) 3
-6 (-5-λ)

To find that the eigenvalues are -2 and 1, and then that the eigenvectors are

(a
-2a)

and

(b
-b)

but after that I'm completely stuck. I know you get into constants and such but I just can't remember how to continue.
You can write (a, -2a)= a(1, -2) and (b, -b) as b(1, -1). Use those vectors as columns in a matrix:
[tex]A= \begin{bmatrix}1 & 1\\ -2 & -1\end{bmatrix}[/tex]
and write [begin]A^{-1}[/itex] as its inverse matrix.
Then
[tex]A^{-1}\begin{bmatrix}1 & 1 \\ -2 & -1\end{bmatrix}A= \begin{bmatrix}-2 & 0\\ 0 & 1\end{bmatrix}[/tex]

Writing your differential equation as differential equation as dY/dt= BY, We can multiply on both sides by [itex]A^{-1}[/itex] and write Y as [itex]Y= AX[/itex] so that \(\displaystyle X= A^{-1}Y\) and the differential equation becomes [itex]d(A^{-1}Y)/dt= dX/dt= A^{-1}BY= (A^{-1}BA)[/itex] which is just
[tex]\frac{dX}{dt}= \begin{bmatrix}-2 & 0 \\ 0 & 1\end{bmatrix}X[/tex]
which is easy to solve. Once you have X, then, of course, Y= AX.
 

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