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MIT18_03S10_ls2

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

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± ± ± ± ± ± ± ± ± 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 first 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 first 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

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

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