lecture_23 (dragged) 1

# lecture_23 (dragged) 1 - 2 MA 36600 LECTURE NOTES WEDNESDAY...

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11 This is a linear diﬀerential equation if we can express ￿ ￿ n−1 ￿ G t, y, y (1) , . . . , y (n−1) = g (t) − p1 (t) y (n−1) − · · · − pn−1 (t) y ￿ − pn (t) y = g (t) − That is, if the diﬀerential equation is in the form j =0 pn−j (t) y (j ) . y (n) + p1 (t) y (n−1) + · · · + pn−1 (t) y ￿ + pn (t) = g (t). We remark equations in the following form are also linear: dn y dn−1 y dy P0 (t) n + P1 (t) n−1 + · · · + Pn−1 (t) + Pn (t) y = G(t) dt dt dt =⇒ n ￿ j =0 Pn−j (t) y (j ) = G(t). Our goal is to show that if we can ﬁnd (n + 1) functions y1 = y1 (t), . . . , yn (t) and Y = Y (t) such that (1) Each yi = yi (t) is a solution to the homogeneous equation n ￿ j =0 (j ) Pn−j (t) yi =0 for i = 1, 2, . . . , n. (2) The n × n determinant is nonzero: ￿ ￿ y1 (t) y2 (t) ··· ￿ ￿ (1) (1) ￿ y (t) y2 (t) ··· ￿1 ￿ ￿ W y1 , y2 , . . . , yn (t) = ￿ ￿ . . .. . . ￿ . . . ￿ ￿ (n−1) (n−1) ￿y (t) y2 (t) · · · 1 (3) Y = Y (t) is a solution to the nonhomogeneous equation n ￿ j =0 ￿ yn (t) ￿ ￿ ￿ (1) yn (t) ￿ ￿ ￿ ￿ (j −1) ￿ ￿ ￿ = ￿y ￿ ￿= 0. ￿ i . . ￿ . ￿ ￿ (n−1) yn (t)￿ Pn−j (t) Y (j ) = G(t). Then the general solution to the diﬀerential equation is ￿n ￿ ￿ y (t) = ci yi (t) + Y (t). i=1 Linear Operators. Given the linear diﬀerential equation P0 (t) dn y dn−1 y dy + P1 (t) n−1 + · · · + Pn−1 (t) + Pn (t) y = G(t) n dt dt dt =⇒ Deﬁne the operator L[y ] = n ￿ i=1 Pn−j (t) y (j ) =⇒ n ￿ i=1 Pn−j (t) y (j ) = G(t). L[y ] = G(t). We will show that this is a linear operator. Given functions f = f (t) and g = g (t) as well as constants c1 and c2 , we have ￿ ￿￿ ￿ ￿ n n ￿ ￿￿ dj dj f dj g L c1 f + c2 g = Pn−j (t) j c1 f (t) + c2 g (t) = Pn−j (t) c1 j + c2 j dt dt dt i=1 i=1 ￿n ￿ ￿n ￿ ￿ ￿ dj f dj g = c1 Pn−j (t) j + c2 Pn−j (t) j dt dt i=1 i=1 = c1 L[f ] + c2 L[g ]. In particular, if {y1 , y2 , . . . , ym } is a set of functions satisfying L[yi ] = 0, then given the linear combination y (t) = c1 y1 (t) + c2 y2 (t) + · · · + cm ym (t) ...
View Full Document

## This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

Ask a homework question - tutors are online