{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

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)

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

View Full Document
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).
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 .

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

View Full Document
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 .
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 ) .

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

View Full Document
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.
*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.

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

View Full Document
8 Problem 3.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}