HW4SOL-s09-correct

# 3 11 y y0 the characteristic equation is and the roots

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Unformatted text preview: he complex number in polar form, 1 e π , where may be any integer. Thus, π e . Setting 0,1,2,3 successively, we obtain the three roots as 1 1 1 1 1 #15. y The characteristic equation is λ , and the roots are given by λ 1 / magnitude R 1 polar angle θ ; cos θ √ √ θ √2e π 5π 4 e cos π 2 1 sin π 2 sin π sin 3π 2 e π eπ e π cos π cos 3π 2 y 0 λ 0 Writing the complex number in polar form, 1 e π , where may be any integer. Thus, π π 1 e . Setting 0,1,2,3,4,5 successively, we obtain the three roots as 1 λ,,,,, , √3 2 , which give us the bases of the general solution y y y y y y c′ e c′ e c′ e c′ e c′ e c e ′ √ e e √ √ A cos A cos √ t 2 t 2 t 2 t 2 B sin B sin t 2 t 2 t 2 t 2 √ √ e e A cos t A cos t A cos A cos B sin B sin √ √ B sin t B sin t Therefore, the general solution is y e √ c cos t 2 c sin t 2 e √ c cos t 2 c sin t 2 c cos t c sin t Chapter 4.3 #11. y ′′′ y0 The characteristic equation is λ , and the roots are given by λ Hence, the homogeneous solution is y c ce ce Bt Ct and substitute those into the ,, 3y ′′ 1 , y ′ 0 2y ′ t e 1 ′′ , y 0 4 3 2 3λ 2λ 0 0,1,2 To get the particular solution, set Y differential equation. Y Y From the equations above, A ′′′ ′′′ Ate , Y ′′ ′′ ′ ′ 3Y 3Y 2Y 2Y e t 3 4 . Therefore, the general solution is 1, B 1 4 , C y using the initial conditions, c c ce 1, c y 1 ce c te 0. t te t 4 3t 3t 4 #12. y 2y ′′′ y ′′ y0 3 , y ′ 0 8y ′ 12y 0 , y ′′ 0 12 sin t 1 , y ′′ 0 2 The characteristic equation is λ , and the roots are given by λ ,,, 2λ λ 8λ 12 0 1, 3, 2 Hence, the homogeneous solution is y ce ce Ae 8Y ′ c cos 2t Bcos t c sin 2t Csin t and substitute those into the To get the particular solution, set Y differential equation. Y ′′′′ 2Y ′′′ Y ′′ 12Y 2 12 sin t 4 e 5 . Therefore, the general solution 2 cos t 5 49 4 sin t 5 From the equation above, A is y ce ce 1 20 , B 5 , C 1 e 20 77 c cos 2t c sin 2t using the initial conditions, c y 81 e 40 81 73 e 520 40 , c 73 520 , c 65 , c 1 e 20 130 4 sin t 5 77 cos 2t 65 49 sin 2t 130 2 cos t 5 Chapter 4.4 #2. The characteristic equation is λ , and the roots are given by λ Hence, the homogeneous solution is y c ce ce 2. Now compute the three determinants e e e 0 0 1 e e e e e e e e e 0 0 1 e 2 ,, y ′′′ y′ t λ 0 0, 1 The Wronskian is evaluated as W 1, e , e Wt 0 0 1 1 0 0 1 0 0 Wt Wt e The solution of the system of equations is u′ t u′ t u′ t tW t Wt tW t Wt tW t Wt t t t u t e u t 2 e u t 2 t 2 e et 2 t1 2 1 Hence the particular solution is Y y u y u y u t 2 t 2 1 t 2 1 t 2 1 The constant is a part of the homogeneous solution, therefore, the general solution is y #3. c ce ce t 2 y ′′′ 2y ′′ y′ 2y e The characteristic equation is λ , and the roots are given by λ Hence, the homogeneous solution is y ce ce ce 6e . Now compute the three determinants e 2e 4e 0 0 1 e e e e 2e 4e 0 0 1 e ,, 2λ λ 2 0 1, 2 The Wronskian is evaluated as W e , e , e Wt 0 0 1 e e e e e e e e e Wt 3e Wt 2 The solution of the system of equations is u′ t u′ t u′ t eW t Wt eW t Wt eW t Wt e u t 3 1 u t 2 e u t 3 e 6 1 t 2 e 3 Hence the particular solution is Y y u y u y u e e 6 e 1 t 2 e 3 1 e 2 1 te 2 The first term is included in homogeneous solutions. Therefore, the general solution is y c e ce ce 1 te 2...
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## This note was uploaded on 10/12/2009 for the course MAE 210792200 taught by Professor Youssef during the Spring '09 term at UCLA.

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