2130_Chapter7_Notes

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

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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...
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