# ode219week5 - Math 219 Lecture 9 Ali Devin Sezer Contents 1...

This preview shows pages 1–3. Sign up to view the full content.

Math 219, Lecture 9 Ali Devin Sezer October 26, 2010 Contents 1 Solution of linear differential equations 1 1.1 Another Solution of (1) . . . . . . . . . . . . . . . . . . . . . . . 2 2 Linear transformations of R 3 3 Linear transformations of R 2 3 1 Solution of linear differential equations Let A be the 2 × 2 matrix A = parenleftbigg a 1 , 1 a 1 , 2 a 2 , 1 a 2 , 2 parenrightbigg and consider the differential equation dx dt = Ax. (1) The goal is to find all functions x : R R 2 satisfying this equation. Use the simplest method you know: Euler’s method. Approximate dx dt with 1 h ( x ( t + h ) - x ( t )) and rewrite (1) as 1 h ( x ( t + h ) - x ( t )) = Ax ( t ) . This gives x ( t + h ) = x ( t ) + hAx ( t ) = [ I + hA ] x ( t ) , (2) where I = parenleftbigg 1 0 0 1 parenrightbigg is the identity matrix. For t = 0 (2) gives x ( h ) = ( I + hA ) x (0) . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
For t = h we get: x (2 h )( I + hA ) x ( h ) = ( I + hA )( I + hA ) x (0) = ( I + hA ) 2 x (0) . Repeating this process n times gives: x ( nh ) = ( I + hA )( I + hA )( I + hA ) · · · ( I + hA ) x (0) = ( I + hA ) n x (0) (3) Thus, “solving” (1) is simply applying the linear transformation ( I + hA ) to the point x (0) over and over again. Then it is clear that in order to understand the solutions of (1) one needs to understand the linear transformations of R 2 . This we take up in Section 3.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern