Applied Differential Equations - Answers

Question 1. [8 points] Determine whether the given set of fuctions is linearly independent on the interval (−∞, ∞). (a) [4 points] f 1 (x) = x, f 2 (x) = x 2 , f 3 (x) = 4x − 3x 2 Solution: Since f 3 (x) = 4 · f 1 (x) − 3 · f 2 (x) =⇒ 4 · f 1 (x) − 3 · f 2 (x) − f 3 (x) = 0, the set of functions f 1 (x), f 2 (x), f 3 (x) is linearly dependent. (b) [4 points] f 1 (x) = 1 + x, f 2 (x) = x, f 3 (x) = x 2 Solution: Let c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) = 0 =⇒ c 1 (1 + x) + c 2 x + c 3 x 2 = 0 =⇒ c 1 + (c 1 + c 2 )x + c 3 x 2 = 0, then c 1 = 0, c 1 + c 2 = 0, c 3 = 0, =⇒ c 1 = c 2 = c 3 = 0. Hence the set of functions f 1 (x), f 2 (x), f 3 (x) is linearly independent. 2 Question 2. [24 points] Consider the differential equation x 3 y ��� + x 2 y �� − 2x y � + 2y = 0 (a) [15 points] Verify that the fuctions of y 1 = x, y 2 = x 2 , y 3 = differential equation. 1 x are solutions of the given Solution: • Check y 1 is the sol of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0 y 1 = x =⇒ y 1� = 1, y 1�� = 0, y 1��� = 0 Then x 3 y 1��� + x 2 y 1�� − 2x y 1� + 2y 1 = 0 + 0 − 2x + 2x = 0. • Check y 2 is the solution of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0 y 2 = x 2 =⇒ y 2� = 2x, y 2�� = 2, y 2��� = 0 Then x 3 y 2��� + x 2 y 2�� − 2x y 2� + 2y 2 = 0 + x 2 · 2 − 2x · 2x + 2x 2 = 0. • Check y 3 is the solution of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0 y3 = 1 =⇒ y 3� = −x −2 , y 3�� = 2x −3 , y 3��� = −6x −4 x Then x 3 y 3��� + x 2 y 3�� − 2x y 3� + 2y 3 = 0. (b) [7 points] Verify that y 1 , y 2 , y 3 in part (a) form a fundamental set of solutions for the given differential equaiton on the interval of (0, ∞). Solution: � � y1 � W (y 1 , y 2 , y 3 )(x) = �� y 1� � y �� 1 y2 y 2� y 2�� y3 y 3� y 3�� � � 1 � � x x2 x � � � = � 1 2x −x −2 � � � � 0 2 2x −3 Then y 1 , y 2 , y 3 form a fundamental set of solutions. � � � � = 6 1 �= 0, on (0, ∞) � x � (c) [2 points] Using y 1 , y 2 and y 3 in part (a), form the general solution of the given differential equation. Solution: The general solution is y(x) = c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 x + c 2 x 2 + c 3 x1 . 3 Question 3. [24 points] Find the general solution of the given second-order differential equation. (a) [8 points] y �� − 7y � + 12y = 0 Solution: m 2 − 7m + 12 = 0 =⇒ (m − 3)(m − 4) = 0 =⇒ m 1 = 3, m 2 = 4 =⇒ y = c 1 e 3x + c 2 e 4x . (b) [8 points] y �� − 6y � + 9y = 0 Solution: m 2 − 6m + 9 = 0 =⇒ (m − 3)2 = 0 =⇒ m 1 = m 2 = 3 =⇒ y = c 1 e 3x + c 2 xe 3x . (c) [8 points] y �� − 4y � + 5y = 0 Solution: m 2 − 4m + 5 = 0 =⇒ m 1,2 = 2 ± i =⇒ y = e 2x (c 1 cos x + c 2 sin x). 4 Question 4. [12 points] Find the general solution of the given third-order differential equation d 3x d 2x + − 2x = 0 dt3 dt2 Solution: m 3 + m 2 − 2 = 0 =⇒ m = 1is one of the roots, then using Long Division, we have m 3 + m 2 − 2 = (m − 1)(m 2 + 2m + 2) Then m 2,3 = −1 ± i . Hence, the general solution is x = c 1 e t + e −t (c 1 cos t + c 2 sin t ) 5 Question 5. [30 points] Solve the given differential equation by undetermined coefficients, y �� + 2y � = 2x + 5 − e −2x subject to the initial conditions y(0) = 0, y � (0) = 12 . Solution: The associated homogeneous equation is y �� + 2y � = 0. Then the auxiliary equation is m 2 + 2m = 0 =⇒ m = 0andm = 2. Hence, the complementary function is y c = c 1 e −2x + c 2 . Now g (x) = 2x + 5 − e −2x =⇒ y p (x) = Ax + B +C e −2x , however, there is duplications between y c and y p . Then the correct assumed particular solution is y p (x) = Ax 2 + B x +C xe −2x . Then y p� (x) = 2Ax + B +C e −2x − 2C xe −2x and y p�� (x) = 2A − 4C e −2x + 4C xe −2x . Hence, y p�� (x) + 2y p� (x) = 2A + B + 4Ax − 2C e −2x = 2x + 5 − e −2x =⇒ 2A + 2B = 5, 4A = 2, −2C = −1 =⇒ A = 12 , B = 2,C = 12 =⇒ y p (x) = 12 x 2 + 2x + 21 xe −2x Then the general solution is 1 1 y = c 1 e −2x + c 2 + x 2 + 2x + xe −2x 2 2 Using I.C. =⇒ 0 = c 1 + c 2 , 12 = −2c 1 + 2 + 12 =⇒ c 1 = 1, c 2 = −1. Therefore, the solution is 1 1 y = e −2x − 1 + x 2 + 2x + xe −2x 2 2 6 Question 6. [30 points] Solve the given differential equation by variation of parameters 3y �� − 6y � + 6y = e x sec x Solution: 1. For 3y �� − 6y � + 6y = 0 we have 3m 2 − 6m + 6 = 0 ⇒ 3(m 2 − 2m + 2) = 0. So � 2 ± 4 − 4 · 2 2 ± 2i m1 , m2 = = = 1±i 2 2 and α = 1 = β. So, y c = e x (c 1 cosx + c 2 sin x). Hence y 1 = e x cos x and y 2 = e x sin x. 2. We have � � y1 w(y 1 , y 2 ) = �� � y1 � � y 2 �� �� e x cos x e x sin x � �=� x x x y2 e cos x − e sin x e sin x + e x cos x � � � � = e x cos x(e x sin x + e x cos x) − e x sin x(e x cos x − e x sin x) = e 2x cos x sin x + e 2x cos2 x − e 2x sin x cos x + e 2x sin2 x = e 2x (cos2 x + sin2 x) = e 2x . 3. 3y �� − 6y � + 6y = e x sec x ⇒ y �� − 2y � + 2y = 13 e x sec x. f (x) = 13 e x sec x. 4. We have So � � 0 w 1 = �� f (x) � � y1 w 2 = �� � y1 � � � � 0 e x sin x y 2 �� �� � = − 1 e 2x sin x sec x � �=� 1 x x x � y2 e sec x e sin x + e cos x 3 3 � � � x � � � e cos x 0 0 � � � = 1 e 2x cos x sec x. =� x 1 x x � f (x) e cos x − e sin x e sec x � 3 3 1 2x w 1 − 3 e sin x sec x 1 1 1 1 = = = − sin x sec x = − sin x = − tan x. 2x w e 3 3 cos x 3 1 2x e cos x sec x w2 3 1 1 u 2� = = = cos x sec x = . 2x w e 3 3 u 1� Therefore, � � � � 1 1 sin x 1 sin x 1 1 1 u 1 = − tan xd x = − dx = − dx = d (cos x) = ln | cos x|. 3 cos x 3 cos x 3 cos x 3 � 3 1 1 u2 = d x = x. 3 3 5. y p = u 1 y 1 + u 2 y 2 = 13 e x cos x ln | cos x| + 13 xe x sin x. 6. y = e x (c 1 cosx + c 2 sin x) + 13 e x cos x ln | cos x| + 13 xe x sin x. 7 Question 7. [20 pionts] A 100-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the capacitance is 10−4 farad. (a) [16 points] Find the charge q(t ) on the capacitor if q(0) = 0. dq Solution: Assume that R d t + C1 q = E (t ) where R = 200 ohms, C = 10−4 farad, and E (t ) = 100. Then dq dq dq 1 200 d t + 104 q = 100 =⇒ 2 d t + 100q = 1 =⇒ d t + 50q = 12 =⇒ q = 100 + ce −50t 1 Using I.C., we get c = − 100 . Hence the solution is q= 1 1 −50t − e . 100 100 (b) [4 points] Find the current i (t ). Solution: i = dq dt 1 = − 100 · (−50)e −50t = 12 e −50t . 8 Question 8. [BONUS: 24 points] Find the general solution of x 4 y �� + x 3 y � − 4x 2 y = 1 given that y 1 = x 2 is a solution of the associated homogeneous equation. [HINT: Reduction of Order & Variation of Parameters] Solution: We are given that y 1 = x 2 is a sol. of x 4 y �� + x 3 y � − 4x 2 y = 1. To find a second solution, we use reduction of order. Let y = x 2 u(x). Then the product rule gives y � = x 2 u � + 2xu and y �� = x 2 u �� + 4xu � + 2n. So x 4 y �� + x 3 y � − 4x 2 y = x 5 (xu �� + 5u � ) = 0 Letting w = u � , this becomes xw � + 5w = 0 Separating variables and integrating we have dw 5 = − d x and ln |w| = −5 ln x +C dt x Thus, w = x −5 and u = − 14 x −4 . A second solution is then y 2 = x 2 · x −4 = x12 , and the general solution of the homogeneous DE is y c = c 1 x 2 + xc22 . To find a particular solution, y p , we use variation of parameters. The Wronskian is W = − x4 . Identifying f (x) = x14 , we obtain 1 −4 x u 1� = 14 x −5 =⇒ u 1 = − 16 1 −1 1 � u 2 = − 4 x =⇒ u 2 = − 4 ln x 1 −4 2 1 −2 So y p = − 16 x x − 14 (ln x)x −2 = − 16 x − 14 x −2 ln x. The general solution is c2 1 1 y = c 1 x 2 + 2 − x 2 − 2 ln x. x 16 4x 9