MIT18_03S10_ls2

MIT18_03S10_ls2 - LS.2 Homogeneous Linear Systems with...

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

View Full Document Right Arrow Icon
± ± ± ± ± ± ± ± ± LS.2 Homogeneous Linear Systems with Constant Coefficients 1. Using matrices to solve linear systems. The naive way to solve a linear system of ODE’s with constant coefficients is by elimi- nating variables, so as to change it into a single higher-order equation. For instance, if x = x + 3 y (1) y = x y we can eliminate x by solving the second equation for x , getting x = y + y , then replacing x everywhere by y + y in the ±rst equation. This gives y 4 y = 0 ; the characteristic equation is ( r 2)( r + 2) = 0, so the general solution for y is 2 t 2 t y = c 1 e + c 2 e . From this we get x from the equation x = y + y originally used to eliminate x ; the whole solution to the system is then 2 t x = 3 c 1 e 2 t c 2 e (2) 2 t 2 t y = c 1 e + c 2 e . We now want to introduce linear algebra and matrices into the study of systems like the one above. Our ±rst task is to see how the above equations look when written using matrices and matrix multiplication. When we do this, the system (1) and its general solution (2) take the forms x 1 3 x = (4) y 1 1 y , 2 t 3 c 1 e 2 t c 2 e 1 2 t (5) x = c 1 e 2 t + c 2 e 2 t = c 1 3 e 2 t + c 2 e . y 1 1 Study the above until it is clear to you how the matrices and column vectors are being used to write the system (1) and its solution (2). Note that when we multiply the column vectors by scalars or scalar functions, it does not matter whether we write them behind or in front of the column vector; the way it is written above on the right of (5) is the one usually used, since it is easiest to read and interpret. We are now going to show a new method of solving the system (1), which makes use of the matrix form (4) for writing it. We begin by noting from (5) that two particular solutions to the system (4) are 3 2 t (6) e and 1 e 2 t . 1 1 6
Background image of page 1

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

View Full DocumentRight Arrow Icon
± ± ± ± ± ± ± ± ² ² ² 7 LS.2 HOMOGENEOUS LINEAR SYSTEMS Based on this, our new method is to look for solutions to (4) of the form x (7) = a 1 e t , y a 2 where a 1 , a 2 and are unknown constants. We substitute (7) into the system (4) to deter- mine what these unknown constants should be. This gives a 1 a 1 t (8) e t = 1 3 e . a 2 1 1 a 2 We can cancel the factor e t from both sides, getting 3 a 1 (9) a 1 = 1 . a 2 1 1 a 2 We have to solve the matrix equation (9) for the three constants. It is not very clear how to do this. When faced with equations in unfamiliar notation, a reasonable strategy is to rewrite them in more familiar notation. If we try this, (9) becomes the pair of equations a 1 = a 1 + 3 a 2 (10) a 2 = a 1 a 2 . Technically speaking, these are a pair of non-linear equations in three variables. The trick in solving them is to look at them as a pair of linear equations in the unknowns a i , with viewed as a parameter. If we think of them this way, it immediately suggests writing them in standard form (1 ) a 1 + 3 a 2 = 0 (11) a 1 + ( 1 ) a 2 = 0 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

MIT18_03S10_ls2 - LS.2 Homogeneous Linear Systems with...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online