Lecture 2 7 , § 5 . 7

Today w e will apply linear algebra t o solve some simple types of differentialequations.

Linear differential equation 8 Let x ( t ) be a function of the variable t . A linear

differential equation i s of t he form× ' ( t ) = a x ( t ) , for some scalar constant a .

-

derivative

Now suppose w e have multiple functions × , ( t ) , x . ( t ) , . . . , x n ( t ) . A system

of linear differential equations i s of the form

xp(t) = a , , x . ( t ) t o o o t a , n x n l t ) .X I ( t ) = as, × ,( t ) t . . . t a n Xnlt) .

u

:

g i l t ) = a n , x . ( t ) t . . . t ann x n ( t ) .

I f I t h i s t h e vector function I N = (§§§.)

and

I 'CH = x .'CH{×,.,,,)

, thegnomthea,

above system c a n b e written i n matrix

:

Xn'( t )

I ' LH = A I C H , where A = (a;j).Goal : Given a vector linear differential equation I ' I t t = A I t t ) , solve fort h e vector-valued function E l t ) .

I n i t i a l v a l u e problem 8 Given a linear system of differential equationsI 'Lt ) = A I C H

w e often want t o find E t t l such tha t t h e ' i n i t i a l value' ECO) i s s om e

fixed vector J , i e . , I 101=5. T h e goal i s then t o construct t h e

unique function I l t t such that # ' ( t ) = A E t t l and Eco) = I .This i s called t h e in i t i a l value problem.

Example : Suppose A = I} j). Then l o v e the linear system of

differential equations I ' I t t = A I C H such that I (o) = (z).Solu t ion 8 S t e p F i n d t h e eigenvalues of A .

de t ( A - 2 I - I = d e t (7,-A j!) = H-H (3-1+3= 2 1 - 1 0 2 t 22+3 = 22 - 1 02 t 24

= (X-6) (A-4) .

T h e eigenvalues a r e t h e roots of d e t ( A -X I . ) . oo. A has eigenvalues7 = 4 , 6 and these a r e a l l possible eigenvalues.

¥ 2 : F i n d ANY nonzero eigenvector i n Ec, and Eg.

Eigenvector Ti, i n F-↳ : F i n d a nonzero vector i n Nut ( A - 4 I z ) .

A-HI-=P;" I.a) = (3, I ] ¥ , ¥ . Yo "o]'§ Linear system

3×1 - X 2 = 0 ⇒ X , = § .% T o get T , take a = 3 and get i t = (b).

Eigenvector I I i n E , 8 A - 6 I s =(73-6 j!,]=/! I ]

1,1221-7122-313,

× , - × , s ochinerlystem ( lo -lo)

I× , = X z

oo. T o get I s , c a n take × , = L , giving a s I , = (!).Step 3 : Since eigenvectors with distinct eigenvalues a r e linearly independent,Jessee that t h e pm?}, ;)3

i s invertible, a n d P "= dettp ( I -I]= I f l - l )2 - 3 I

= [ " 2 " 2 ].312 -112

Let E .µ =P ' ' I Col = [ " i "2) (?) = 3 [ "2) t-f".)312 -112 312 -112

=/-"Itf') --fi:).9 1 , - 1

Then t h e general solution I l t ) satisfying 5101 = (z) i s given by

I c t ) = c , I e ' t t c , i i . e ' t = I (1)eat + I f ;] e ' t2 3 2

=/.".EE/t/%ee:tI=fte4t-tzeotj.- 116 712 fe l t t7ze6t

T h e method for t h e 2 × 2 mat r i x c a n be generalized t o n x n matr ix Aa s long a s I t has i n distinct eigenvalues.

Solving R 'Lt) = A I t t ) with i n i t ia l va lue I l o ) = I 8-

Warning : Method only applies i f A i s diagonalizable, o r equivalently,

if A has n distinct eigenvalues.

Step1 : F i n d t h e n distinct eigenvalues X, , . . . , I n for A by finding t h eroots of d e t (A-XIN).

Step 2 : F i n d a nonzero eigenvector I i i n each eigenspace Ex;, E l , . . . , n .

Step 3 : F i n d P " for t h e matrix P = [ I I I . . . I n ] ,where t h e

I . a r e a s i n step 2 .

step 4 : Compute I =/}:)

= P" I l o ) = P " 5 ' .

Step 5 : T h e general solution with initial va lue I l o ) = IT i s given by

I t t ) = C, I , c ' ' t t c ,#et-t t . . . t chine a t .