2130_Chapter7_Notes

# 2 1 general solution x1 c1 e2t c2 et x2 2 c1 e2t

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: e 2) x′ = 3x1 − 2x2 1 x′ = 2x1 − 2x2 2 =⇒ x′′ − x′ − 2x1 = 0 1 1 Characteristic equation: r 2 − r − 2 = 0 =⇒ (r − 2)(r + 1) = 0 =⇒ r = 2, −1 =⇒ x1 = c1 e2t + c2 e−t =⇒ x′ = 2c1 e2t − c2 e−t , 1 x2 = = 1 1 ′ 2t −t 2 (3x1 − x1 ) = 2 3(c1 e + c2 e ) − 1 2t −t − 2c e2t + c e−t 1 2 2 3c1 e + 3c2 e (2c1 e2t and − c2 e−t ) = 1 c1 e2t + 2c2 e−t . 2 1 General solution: x1 = c1 e2t + c2 e−t , x2 = 2 c1 e2t + 2c2 e−t . 7/9 4 Exa 2 (slide 3) General solution: x1 = c1 e2t + c2 e−t , x2 = 1 c1 e2t + 2c2 e−t . 2 Use the initial conditions to solve for c1 , c2 : x1 (0) = 3 x2 (0) = 1 2 c1 1 2 c1 =⇒ + c2 = 3 + 2c2 = 1 2 =⇒ −c1 − c2 = −3 c1 + 4c2 = 1 2 =⇒ 3c2 = −2 =⇒ c2 = − 3 =⇒ c1 = =⇒ 9 3 2 − −3 =⇒ c1 = 11 3 1 x1 = 3 (11e2t − 2e−t ), x2 = 1 (11e2t − 8e−t ). 6 8/9 Matrix Equations x′ = 3x1 − 2x2 , 1 x′ = 2x1 − 2x2 , 2 x1 (0) = 3 x2 (0) = 1/2 There is a better way to approach this problem, involving...
View Full Document

Ask a homework question - tutors are online