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# 285lect17 - 1 Lecture 17 Denition 1 A set of solutions y1(x...

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1 Lecture 17 De°nition 1 A set of solutions y 1 ( x ) ; y 2 ( x ) ; :::; y n ( x ) ; a < x < b , of the linear ODE of order n y ( n ) + p 1 ( x ) y ( n ° 1) + p 2 ( x ) y ( n ° 2) + ::: + p n ( x ) y ( x ) = 0 is called the fundamental set of solutions if W ( y 1 ; y 2 ; :::; y n ) 6 = 0 at least at one point x satisfying a < x < b: Theorem 2 IF y 1 ( x ) ; y 2 ( x ) ; :::; y n ( x ) ; a < x < b , is a fundamental set of solutions of the linear ODE of order n y ( n ) + p 1 ( x ) y ( n ° 1) + p 2 ( x ) y ( n ° 2) + ::: + p n ( x ) y ( x ) = 0 ; then, every solution y ( x ) can be written as a linear combination of solutions y 1 ( x ) , y 2 ( x ) ; ::; y n ( x ) : y ( x ) = C 1 y 1 ( x ) + C 2 y 2 ( x ) + ::: + C n y n ( x ) ; a < x < b . If y 1 ( x ) ; y 2 ( x ) ; :::; y n ( x ) ; a < x < b; is a fundamental set of solutions, then y ( x ) = C 1 y 1 ( x ) + C 2 y 2 ( x ) + ::: + C n y n ( x ) ; a < x < b; is called general solution of the linear ODE of order n . Example 3 Consider the following initial value problem: y 000 ° 6 y 00 + 11 y 0 ° 6 y = 0; y (0) = 0 ; y 0 (0) = 0 and y 00 (0) = 3 : First, we need to °nd the fundamental set of solutions. Let y ( x ) = e rx . Then y 000 ° 6 y 00 + 11 y 0 ° 6 y = e rx ° r 3 ° 6 r 2 + +11 r ° 6 ± = 0 ) r 3 ° 6 r 2 + 11 r ° 6 = 0 : We know that r 1 r 2 r 3 = ° 6 .

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