M343 Homework 6. Enrique Areyan May 31, 2013

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1 M343 Homework 6 Enrique Areyan May 31, 013 Section y + 5y = 3sin(t). The general solution is given by: y h : Characteristic equation: r + r + 5 = 0 r = 1 ± i. The solution in this case is: y h = C 1 e t cos(t) + C e t sin(t) y p : Guess y p = Acos(t) + Bsin(t). Then, y p = Asin(t) + Bcos(t) and p = 4Acos(t) 4Bsin(t). 3sin(t) = p + y p + 5y p = 4Acos(t) 4Bsin(t) + ( Asin(t) + Bcos(t)) + 5(Acos(t) + Bsin(t)) = (A + 4B)cos(t) + ( 4A + B)sin(t) A + 4B = 0 = A = 4B = A = B + B = 3 = 17B = 3 = B = 3 17 y = C 1 e t cos(t) + C e t sin(t) 1 17 cos(t) sin(t) 3. y 3y = 3te t. The general solution is given by: y h : Characteristic equation: r r 3 = 0 = 0 r 1 = 3; r = 1. The solution in this case is: y h = C 1 e 3t + C e t y p : Guess y p = t(at + B)e t y p = e t (At + Bt). Then: 3te t = p y p 3y p y p = e t [ At + (A B)t + B] p = e t [At + ( 4A + B)t + A B] = e t [At + ( 4A + B)t + A B] (e t [ At + (A B)t + B]) 3(e t (At + Bt)) = e t [t (A + A 3A) + t( 4A + B 4A + B 3B) + (A B B)] = e t [t( 8A) + A 4B] 1

2 8A = 3 = A = 3 8 A 4B = 0 = A = B = B = 3 16 ( 3 y = C 1 e 3t + C e t + 8 t + 3 ) 16 t e t y + 4y = e t e t. The general solution is given by: y h : Characteristic equation: r + r + 4 = 0 r = 1 ± 15i. The solution in this case is: ( ) ( ) y h = C 1 e t/ cos t + C e t/ sin t y p : Guess y p = Ae t + Be t. Then, y p = Ae t Be t and p = Ae t + Be t. e t e t = p + y p + 4y p = Ae t + Be t + Ae t Be t + 4Ae t + 4Be t = 6Ae t + 4Be t 6A = 1 = A = 1 6 y = C 1 e t/ cos 4B = 1 = B = 1 4 ( ) ( ) t + C e t/ sin t + et 6 e t y y = t, y(0) = 0; y (0) = 1. The general solution is given by: y h : Characteristic equation: r + r = 0 = 0 r 1 = ; r = 1. The solution in this case is: y p : Guess y p = At + B. Then, y p = A and p = 0. y h = C 1 e t + C e t t = + y y = 0 + A (At + B) = At + A B

3 A = = A = 1 Now, solve for C 1, C : A B = 0 = 1 B = 0 = B = 1 y = C 1 e t + C e t t 1 y(0) = 0 = C 1 + C 1 = C 1 + C = 1 = C 1 = 1 The solution to the I.V.P is: y (0) = 1 = C 1 + C 1 = C 1 + C = = C = 1 y = 1 e t + e t t y + y = te t + 4, y(0) = 1; y (0) = 1. The general solution is given by: y h : Characteristic equation: r r + 1 = 0 = 0 (r 1) = 0. The solution in this case is: y h = C 1 e t + C te t y p : Guess y p = t (At + B)e t + C y p = (At 3 + Bt )e t + C. Then: y p = e t [At 3 + (3A + B)t + Bt] p = e t [At 3 + (6A + B)t + (6A + 4B)t + B] te t + 4 = p y p + y p = e t [At 3 + (6A + B)t + (6A + 4B)t + B] (e t [At 3 + (3A + B)t + Bt]) + (At 3 + Bt )e t + C = e t [6At + B] + C 6A = 1 = A = 1 6 B = 0 = B = 0 C = 4 y = C 1 e t + C te t t3 e t + 4 Now, solve for C 1, C : The solution to the I.V.P is: y(0) = 1 = C = C 1 = 3 y (0) = 1 = 3 + C = C = 4 y = 3e t + 4te t t3 e t + 4 3

4 y = 3sin(t), y(0) = ; y (0) = 1. The general solution is given by: y h : Characteristic equation: r + 4 = 0 = 0 r = ±i. The solution in this case is: y p : Guess y p = t[asin(t) + Bcos(t)]. Then: 3sin(t) = p + 4y p y h = C 1 cos(t) + C sin(t) y p = [Asin(t) + Bcos(t)] + t[acos(t) Bsin(t)] p = 4Acos(t) 4Bsin(t) 4tAsin(t) 4Btcos(t) = 4Acos(t) 4Bsin(t) 4tAsin(t) 4Btcos(t) + 4t[Asin(t) + Bcos(t)] = 4Acos(t) 4Bsin(t) 4tAsin(t) 4Btcos(t) + 4Atsin(t) + 4Btcos(t) = 4Acos(t) 4Bsin(t) 4A = 0 = A = 0 4B = 3 = B = 3 4 y = C 1 cos(t) + C sin(t) 3 4 tcos(t) Now, solve for C 1, C : y(0) = = C 1 = C 1 = y (0) = 1 = 3 + C = C = = C = 1 8 The solution to the I.V.P is: y = cos(t) 1 8 sin(t) 3 4 tcos(t) y + 5y = 4e t cos(t), y(0) = 1; y (0) = 0. The general solution is given by: y h : Characteristic equation: r + r + 5 = 0 = 0 r = 1 ± i. The solution in this case is: y p : Guess y p = t[asin(t) + Bcos(t)]e t. Then: y h = C 1 e t cos(t) + C e t sin(t) y p = (e t te t )[Asin(t) + Bcos(t)] + te t [Acos(t) Bsin(t)] p = ( e t +te t )[Asin(t)+Bcos(t)]+(e t te t )[Acos(t) Bsin(t)]+te t [ 4Asin(t) 4Bcos(t)] 4

5 4e t cos(t) = p + y p + 5y p Bt 4B = 0 = B = 0 At + 4A = 4 = A = 1 Now, solve for C 1, C : y = C 1 e t cos(t) + C e t sin(t) + te t sin(t) y(0) = 1 = C 1 = C 1 = 1 y (0) = 0 = 1 + C C = 1 The solution to the I.V.P is: y = e t cos(t) + 1 e t sin(t) + te t sin(t) Section y + y = 16e t/ y y = 4et/. The solution to the homogeneous equation is: r r = 0 (r 1 ) = 0 y h = C 1 e t/ + C te t/ Let y 1 = e t/ and y = te t/. On the one hand, using the method of variation of parameters: e t/ te t/ (y 1, y ) = e t/ e t/ + tet/ = et u 1 = y g = ( tet/ )(4e t/ ) e t = 4t = u 1 = t + C 1 u = y 1 g = (et/ )(4e t/ ) e t = 4 = u = 4t + C y = u 1 y 1 + u y = ( t + C 1 )e t/ + (4t + C )e t/ t General solution: y = C 1 e t/ + C te t/ + t e t/ The particular solution, using the method of variation of parameters is y p = t e t/. On the other hand, using the method of undetermined coefficients: guess the solution y p : y p = At e t/ ; = y p = Ate t/ + A t e t/ ; = p = Ae t/ + Ate t/ + A 4 t e t/ 5

6 The particular solution y p must satisfy the equation: 4e t/ = p y p y p = Ae t/ + Ate t/ + A 4 t e t/ Ate t/ A t e t/ + A 4 t e t/ = e t/ [A + At + A 4 t At A t + A 4 t ] = e t/ [t ( A 4 + A 4 A ) + t(a A) + A] From which we can conclude that A = 4 A =. So the particular solution is y p = t e t/ The two methods agree y = tan(t), 0 < t < π/. The solution to the homogeneous equation is: r + 1 = 0 r = ±i y h = C 1 cos(t) + C sin(t) Let y 1 = cos(t) and y = sin(t). Using the method of variation of parameters: (y 1, y ) = cos(t) sin(t) sin(t) cos(t) = 1 u 1 = y g = sin(t)tan(t) = sin (t) sin (t) = u 1 = 1 cos(t) cos(t) dt = cos(t) sec(t)dt = sin(t) ln(sec(t)+tan(t))+c 1 u = y 1 g = cos(t)tan(t) 1 = sin(t) = u = cos(t) + C y = u 1 y 1 + u y = (sin(t) ln(sec(t) + tan(t)) + C 1 )cos(t) + ( cos(t) + C )sin(t) General solution: y = C 1 cos(t) + C sin(t) ln(sec(t) + tan(t))cos(t) y = 3csc(t), 0 < t < π/. The solution to the homogeneous equation is: r + 4 = 0 r = ±i y h = C 1 cos(t) + C sin(t) Let y 1 = cos(t) and y = sin(t). Using the method of variation of parameters: (y 1, y ) = cos(t) sin(t) sin(t) cos(t) = u = y 1 g u 1 = y g = sin(t)3csc(t) = cos(t)3csc(t) = 3 = u 1 = 3 t + C 1 = 3 cotan(t) = u = 3 4 ln(sin(t)) + C y = u 1 y 1 + u y = ( 3 t + C 1)cos(t) + ( 3 4 ln(sin(t)) + C )sin(t) General solution: y = C 1 cos(t) + C sin(t) 3 tcos(t) ln(sin(t))sin(t) 6

7 10. y + y = et 1 + t The solution to the homogeneous equation is: r r + 1 = 0 (r 1) = 0 y h = C 1 e t + C te t Let y 1 = e t and y = te t. Using the method of variation of parameters: (y 1, y ) = et te t e t e t + te t = e t + te t te t = e t u 1 = y g u = y 1 g = tet et 1 + t e t = t 1 + t = u 1 = 1 ln(1 + t ) + C 1 = et et 1 + t e t = t = u = tan 1 (t) + C y = u 1 y 1 + u y = ( 1 ln(1 + t ) + C 1 )e t + (tan 1 (t) + C )te t General solution: y = C 1 e t + C te t et ln(1 + t ) + te t tan 1 (t) 13. t y = 3t 1, t > 0; y 1 (t) = t ; y (t) = t 1. Before proceeding to solve this by variation of parameters, we need to write this equation in standard form: t y = 3 1 t y 1 and y are solutions of the corresponding homogeneous equation since: 1 t y 1 = = 0 u 1 = y g u = y 1 g t y = t 3 t 3 = 0 (y 1, y ) = t t 1 t t = 1 = 3 t 1 (3 = 3 t (3 = 3 1 t ) 1 t = 1 t 1 3t 3 = u 1 = ln(t) + 1 6t + C 1 = t = u = t3 3 + t 3 + C y = u 1 y 1 + u y = (ln(t) + 1 6t + C 1)t + ( t3 3 + t 3 + C )t 1 Incorporating the quadratic term into the solution of the homogeneous equation: So the particular solution is y = C 1 t + C t 1 + t ln(t) + 1 y p = t ln(t) + 1 7

8 17. x 3xy + 4y = x ln(x), x > 0; y 1 (x) = x ; y (x) = x ln(x). Before proceeding to solve this by variation of parameters, we need to write this equation in standard form: 3 x y + 4 x y = ln(x) y 1 and y are solutions of the corresponding homogeneous equation since: 1 3 x y x y 1 = 3 x x + 4 x x = 6 6 = 0 3 x y + 4 x y = ln(x) x (xln(x) + x) + 4 x x ln(x) = 0 (y 1, y ) = x x ln(x) x xln(x) + x = x3 ln(x) + x 3 x 3 ln(x) = x 3 u 1 = y g u = y 1 g y = u 1 y 1 + u y = ( ln3 (x) 3 So the particular solution is = x ln (x) x 3 = x ln(x) x 3 = ln (x) x = ln(x) x + C 1 )(x ) + ( ln (x) y p = 1 6 x ln 3 (x) = u 1 = ln3 (x) 3 = u = ln (x) + C + C 1 + C )x ln(x) = C 1 x + C x ln(x) x ln 3 (x) Section Consider the equation: + t + t y + t 3 y = ln(t). Since the functions: t, t, t 3, ln(t) are continuous everywhere when t > 0, the interval in which the solution is certain to exists is t > Let f 1 t(t) = t 3, f (t) = t + 1, f 3 (t) = t t. These functions are linearly dependent since: k 1 f 1 (t) + k f (t) + k 3 f 3 (t) = 0 Hold for every t, in particular t = 1, 0, 1 5k 1 + k + 3k 3 = 0 = 14k 1 = 0 3k 1 + k = 0 = k = 3k 1 k 1 + k + k 3 = 0 = k 3 = 5k 1 Hence, k 1 = k = k 3 1. Consider the fourth O.D.E y (4) + = 0 and the functions 1, t, cos(t), sin(t). These are solutions since: 1 (4) + 1 = = 0 t (4) + t = = 0 cos(t) (4) + cos(t) = cos(t) cos(t) = 0 sin(t) (4) + sin(t) = sin(t) sin(t) = 0 The ronskian is: 1 t cos(t) sin(t) (1, t, cos(t), sin(t)) = 0 1 sin(t) cos(t) 1 sin(t) cos(t) 0 0 cos(t) sin(t) = 0 cos(t) sin(t) 0 0 sin(t) cos(t) 0 sin(t) cos(t) = sin (t) + cos (t) = 1 8

9 15. Consider the third O.D.E x = 0 and the functions 1, x, x 3. These are solutions since: x(1) (1) = 0 0 = 0 x(x) (x) = 0 0 = 0 x(x 3 ) (x 3 ) = 6x 6x = 0 The ronskian is: 1 x x 3 (1, x, x 3 ) = 0 1 3x 0 0 6x = 1 3x 0 6x = 6x 9