MTR_500_HW_3 - KEY

MTR_500_HW_3 - KEY - Assignment – 3 Prepared by Ihab Abu...

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Unformatted text preview: Assignment – 3 Prepared by Ihab Abu Younis ay ′′ + by ′ + cy = 0 To find the solution we fund the eigenvalues as follow : aλ2 + bλ + +c = 0 ⇒ λ1, 2 = − b ± b 2 − 4ac 2a for b 2 − 4ac = 0 c= b2 4a −b = 2a −b t 2a λ1, 2 ⇒ so one solution is y1 = e if y 2 = v (t ) y1 = v(t )e 2 a then ay ′′ + by ′ + cy 2 = 0 2 2 −b t −b −b −b −b t t − b 2a t b2 b 2a t a v ′′(t )e 2 a + v ′(t )( e ) + 2 e 2 a v (t ) − e v ′(t ) 2a 2a 4a −b −b −b t t b 2a t + b v ′(t )e 2 a − e v(t ) + c e 2 a v(t ) = 0 2a ⇒ av ′′(t )e 2 a − be 2 a v ′(t ) + −b t −b t t t b 2 2a t b 2 2a t e v(t ) + bv ′(t )e 2 a − e v(t ) + ce 2 a v(t ) = 0 4a 2a −b −b −b −b b2 av ′′(t ) + − 4a + c v(t ) = 0 b2 b2 av ′′(t ) + − 4a + 4a v(t ) = av ′′(t ) = 0 ⇒ v ′′(t ) = 0 ⇒ v ′(t ) = k ⇒ v (t ) = kt ⇒ y 2 = kty1 ⇒ y (t ) = c1e −b t 2a k is an arbitrary constant + c 2 te −b t 2a For the solution to be linearly dependent the the following condition must be true : y1 = contant y2 −b y c e 2a c ⇒ 1 = 1 −b = 1 ≠ constant t y2 c2 t c 2 te 2 a so the two solutions y1 & y 2 are linearly independent t Problem – 2: y ′′ − 2 y ′ + y = te t + 4 To find y h y ′′ − 2 y ′ + y = 0 λ2 − 2λ + 1 = 0 ( λ − 1) 2 = 0 λ1, 2 = 1 y h = c1e t + c 2 te t 1 − Finding y p using undetermined coefficients method : assume that y p = C 2 te t + C1e t + K y ′p = C 2 te t + C 2 e t + C1e t ′ y ′p = 2C 2 e t + C 2 te t + C1e t ′ ⇒ y ′p − 2 y ′p + y p = te t + 4 t t t ⇒ 2C 2 e t + C 2 te t + C1e t − 2 C 2 te t + C 2 e t + C1e t + C 2 te + C1e + K = te + 4 K = te t + 4 ( ) 2 - Finding y p using the variation of parameters : y1 = e t ′ W = y1 y ′ − y1 y 2 2 = e t e t + te t − te 2t = e 2t y p = − y1 ∫ y2 r yr dt + y 2 ∫ 1 dt W W t t te te + 4 e t te t + 4 = −e t ∫ dt + te t ∫ dt e 2t e 2t y 2 = te t ( ) ( ) ( ) = −e t ∫ te −t te t + 4 dt + te t ∫ e −t te t + 4 dt t3 t2 = −e t − 4te −t − 4e −t + te t − 4e −t 3 2 3t 3t 3t te te te =− + 4t + 4 + − 4t = +4 3 2 6 ( ) ( ) Problems 4 – 10: 4- 2 − 5 x′ = x 1 − 2 Solution : 2 − 5 A= 1 − 2 ⇒ A − λI = 0 ⇒ 2−λ 1 −5 = ( 2 − λ )( − 2 − λ ) + 5 = λ2 + 1 = 0 −2−λ ⇒ λ1, 2 = ± j λ1 = j 2 x1 − 5 x 2 = jx1 x1 − 2 x 2 = jx 2 ⇒ x1 = 0.913 x 2 = 0.365 − 0.1826 j λ1 = − j 2 x1 − 5 x 2 = − jx1 x1 − 2 x 2 = − jx 2 ⇒ x1 = 0.913 x 2 = 0.365 + 0.1826 j Center 5- −1 − 1 x′ = x 0 − 0.25 Solution : −1 − 1 A= 0 − 0.25 ⇒ A − λI = 0 ⇒ −1− λ 0 −1 = ( − 1 − λ ) ( − 0.25 − λ ) = 0 − 0.25 − λ ⇒ λ1 = −1 λ 2 = −0.25 λ1 = −1 − x1 − x 2 = − x1 − 0.25 x 2 = − x 2 ⇒ x1 = 1 x2 = 0 λ1 = −0.25 − x1 − x 2 = −0.25 x1 − 0.25 x 2 = −0.25 x 2 ⇒ x1 = −0.8 x 2 = 0.6 Stable Node 6- 3 − 4 x′ = x 1 − 1 Solution : 3 − 4 A= 1 − 1 ⇒ A − λI = 0 ⇒ 3−λ 1 −4 = ( 3 − λ ) ( − 1 − λ ) + 4 = λ2 − 2λ + 1 = 0 −1− λ ( λ − 1) 2 = 0 ⇒ λ1, 2 = 1 Unstable Node 7- 2 − 5 / 2 x′ = x −1 9 / 2 Solution : − 5 / 2 2 A= −1 9 / 2 ⇒ A − λI = 0 ⇒ 2−λ 9/2 −5/ 2 45 37 = ( 2 − λ )( −1− λ ) + = λ2 − λ + =0 −1− λ 4 4 1 ± 1 − 37 1 = ±3j 2 2 ⇒ λ1, 2 = λ1 = 1 −3j 2 x1 = 0.7625 1 +3j 2 x1 = 0.7625 x 2 = 0.4575 − 0.475 j λ2 = x 2 = 0.4575 + 0.475 j Stable Focus 8- 2 − 1 x′ = x 3 − 2 Solution : 2 − 1 A= 3 − 2 ⇒ A − λI = 0 ⇒ 2−λ 3 −1 = ( 2 − λ )( − 2 − λ ) + 3 = λ2 − 1 = 0 2−λ ⇒ λ1, 2 = ±1 λ1 = 1 x1 = 0.7017 λ 2 = −1 x1 = 0.36162 x 2 = 0.7017 x 2 = 0.9487 Saddle 9- 1 3 x′ = x 3 − 1 Solution : 1 3 A= 3 − 1 ⇒ A − λI = 0 ⇒ 1− λ 3 3 = (1 − λ ) ( − 1 − λ ) − 3 = λ 2 − 4 = 0 −1− λ ⇒ λ1, 2 = ±2 λ1 = 2 x1 = 0.4998 λ 2 = −2 x1 = −0.8662 x 2 = −0.8662 x 2 = −0.4998 Saddle 10- 3 − 2 x′ = x 2 − 2 Solution : 3 − 2 A= 2 − 2 ⇒ A − λI = 0 ⇒ 3−λ 2 −2 = ( 3 − λ ) ( − 2 − λ ) + 4 = λ2 − λ − 2 = 0 −2−λ ⇒ λ = ( λ − 2 ) ( λ + 1) λ1 = 2 λ 2 = −1 λ1 = 2 x1 = 0.8944 λ 2 = −1 x1 = 0.4472 x 2 = −0.4472 x 2 = 0.8944 Saddle ...
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