MIT18_03S10_ls6

# MIT18_03S10_ls6 - LS.6 Solution Matrices In the literature...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: LS.6 Solution Matrices In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. You need to get used to the terminology. As before, we state the definitions and results for a 2 × 2 system, but they generalize immediately to n × n systems. 1. Fundamental matrices. We return to the system (1) x = A ( t ) x , with the general solution (2) x = c 1 x 1 ( t ) + c 2 x 2 ( t ) , where x 1 and x 2 are two independent solutions to (1), and c 1 and c 2 are arbitrary constants. We form the matrix whose columns are the solutions x 1 and x 2 : x 1 x 2 (3) X ( t ) = ( x 1 x 2 ) = . y 1 y 2 Since the solutions are linearly independent, we called them in LS.5 a fundamental set of solutions, and therefore we call the matrix in (3) a fundamental matrix for the system (1). Writing the general solution using X(t) . As a first application of X ( t ), we can use it to write the general solution (2) eﬃciently. For according to (2), it is x = c 1 x 1 + c 2 x 2 = x 1 x 2 c 1 , y 1 y 2 y 1 y 2 c 2 which becomes using the fundamental matrix c 1 (4) x = X ( t ) c where c = , ( general solution to (1) ) . c 2 Note that the vector c must be written on the right, even though the c ’s are usually written on the left when they are the coeﬃcients of the solutions x i . Solving the IVP using X(t) . We can now write down the solution to the IVP (5) x = A ( t ) x , x ( t ) = x . Starting from the general solution (4), we have to choose the c so that the initial condition in (6) is satisfied. Substituting t into (5) gives us the matrix equation for c : X ( t ) c = x . Since the determinant | X ( t ) | is the value at t of the Wronskian of x 1 amd x 2 , it is non-zero since the two solutions are linearly independent (Theorem 5.2C). Therefore the inverse matrix exists (by LS.1), and the matrix equation above can be solved for c : c = X ( t ) − 1 x ; using the above value of c in (4), the solution to the IVP (1) can now be written (6) x = X ( t ) X ( t ) − 1 x . 25 26 18.03 NOTES: LS. LINEAR SYSTEMS Note that when the solution is written in this form, it’s “obvious” that x ( t ) = x , i.e., that the initial condition in (5) is satisfied....
View Full Document

## This note was uploaded on 10/21/2011 for the course MATH 18.03 taught by Professor Vogan during the Fall '09 term at MIT.

### Page1 / 7

MIT18_03S10_ls6 - LS.6 Solution Matrices In the literature...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online