{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw7 - Week 7 0 4.1 2nd Order Linear Equations 1 4.2...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Week 7: 0. 4.1 2nd Order Linear Equations 1. 4.2 Constant Coef, Homogeneous DE 2. 4.3 Undetermined Coef 3. 4.5 Springs (Homog case)
Background image of page 1

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

View Full Document Right Arrow Icon
2 General Solution: (1) to solve the 2nd order linear DE y + a 1 y + a 2 y = 0 find two linearly independent solutions y 1 and y 2 , then the general solution is y H = c 1 y 1 + c 2 y 2 . (2) to solve the non-homogeneous 2nd order linear DE y + a 1 y + a 2 y = F, find a particular solution y p , then the general solution is y = y H + y p , where y H is the solution of (1).
Background image of page 2
Problem 1: Determine all values r so y = e rx is a solution to y - 4 y + 3 y = 0 . Find the general solution. Solution: For y = e rx , we have y = re rx and y = r 2 e rx . Substitution in the DE gives 0 = y - 4 y + 3 y = r 2 e rx - 4 re rx + 3 e rx = ( r 2 - 4 r + 3) e rx . Now r 2 - 4 r + 3 = ( r - 3)( r - 1) = 0 has roots r 1 = 1 and r 2 = 3 , so we get solutions y 1 = e x and y 2 = e 3 x , so the general solution is y = c 1 e x + c 2 e 3 x .
Background image of page 3

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

View Full Document Right Arrow Icon
4 Thursday: Constant Coef, Homogeneous DE When the DE y + a 1 y + a 2 y = 0 has coefficients a 1 and a 2 that are constant, the two linearly independent solutions y 1 and y 2 in the general solution y H = c 1 y 1 + c 2 y 2 may be determined using the roots of the characteristic polynomial P ( r ) = r 2 + a 1 r + a 2 .
Background image of page 4
Three cases: (1) For distinct real roots r 1 and r 2 , y = c 1 e r 1 x + c 2 e r 2 x . (2) For a repeated real root r 1 the 2nd independent solution is y 2 = xe r 1 x , and y = c 1 e r 1 x + c 2 xe r 1 x . (3) For a pair of complex roots r = a + bi and r = a - bi with a and b real, b > 0, y = c 1 e ax cos( bx ) + c 2 e ax sin( bx ) .
Background image of page 5

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

View Full Document Right Arrow Icon
6 Tues: Recitation on 4.2. Problem 2: Solve y - 6 y + 34 y = 0 . Solution: The characteristic polynomial is r 2 - 6 r + 34 , so the quadratic formula gives roots 6 ± 6 2 - 4 · 34 2 = 3 ± 5 i. (why? ) So y = e 3 x ( c 1 cos 5 x + c 2 sin 5 x ) is the general solution.
Background image of page 6
*Postscript: We may solve the IVP y + 4 y + 4 y = 0 , y (0) = 1 , y (0) = 4 . We also may solve an IVP for Problem 2 (time permitting). In any case, the calculation of constants c 1 and c 2 to solve an IVP look rather different in (at least) four cases: the three types of roots in the homogeneous DE, and (fourth) the non-homogeneous DE.
Background image of page 7

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

View Full Document Right Arrow Icon
8 Problem 3.
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}