Name: Solutions Exam 2

Transcription

1 Intruction. Anwer each of the quetion on your own paper. Put your name on each page of your paper. Be ure to how your work o that partial credit can be adequately aeed. Credit will not be given for anwer (even correct one) without upporting work. A table of Laplace tranform and the tatement of the main partial fraction decompoition theorem have been appended to the exam. There are 60 point poible. 1. [10 Point] Compute the Laplace tranform of each of the following function. You may ue the attached table, but be ure to identify which formula you are uing by citing the number() or name of the formula in the table. (a) f(t) = e 2t (t 3 co 3t) Solution. f(t) = e 2t t 3 e 2t co 3t o ( ) 6 F () = ( 2) 2 6 = ( + 2) ( + 2) (b) g(t) = 3t in 4t. Solution. Since L {3 in 4t} = 3 4, the Tranform Derivative Principle give G() = d d ( ) ( 1) 2 = = ( ) 2 24 ( ) [8 Point] Find the Laplace tranform Y () of the olution y(t) of the initial value problem. 2y + 5y + 2y = co 2t, y(0) = 3, y (0) = 2. Note that you are aked to find Y (), but not y(t). Solution. Apply the Laplace tranform to the differential equation, uing linearity and the input derivative principle to get Thi give or 2( 2 Y () y(0) y (0)) + 5(Y () y(0)) + 2Y () = Y () Y () Y () = ( )Y () 6 11 = Math 2065 Section 2 October 12,

2 Solve for Y () to get Y () = ( )( 2 + 4). 3. [12 Point] Compute the invere Laplace tranform of each of the following rational function. (a) F () = ( + 1)( + 2)( 3) Solution. Expand F () into partial fraction. product of ditinct linear term, where F () = A B Since the denominator i a C 3, A = ( + 2)( 3) = 1 = 1 4 = 1 4, B = ( + 1)( 3) = = 1, = 2 5 C = ( + 1)( + 2) = = 35 = = 7 4. Thu, (b) G() = Solution G() = f(t) = L 1 {F ()} = 1 4 e 1 + e 2t e3t = ( + 2) (( + 2) 2) + 5 = = ( + 2) = 2( + 2) + 1 ( + 2) ( + 2) ( + 2) ( + 2) Thu L 1 {G()} = 2e 2t co 5t e 2t in 5t. Math 2065 Section 2 October 12,

3 4. [20 Point] Find the general olution of each of the following homogeneou differential equation. (a) 2y + 9y + 10y = 0 Solution. q() = = (2 + 5)( + 2) which ha root 5/2 and 2. Thu, y = c 1 e 5t/2 + c 2 e 2t. (b) y + 6y + 34y = 0 Solution. q() = = ( + 3) o the root are 3 ± 5i. Hence, (c) 9y + 12y + 4y = 0 y = c 1 e 3t co 5t + c 2 e 3t in 5t. Solution. q() = = (3 + 2) 2 which ha a ingle root 2/3 with multiplicity 2. Thu, y = c + 1e 2t/3 + c 2 te 2t/3. (d) y + 9y = 0 Solution. q() = = ( 2 + 9) which ha root 0, ±3i. Thu, y = c 1 + c 2 co 3t + c 3 in 3t. 5. [10 Point] Find the general olution of the following differential equation: You may ue whatever method you prefer. y + 6y + 8y = 5e 4t. Solution. Ue the method of undetermined coefficient. The characteritic polynomial i q() = = ( + 2)( + 4) which ha root 2 and 4. Thu B q = {e 2t, e 4t } and y h = c 1 e 2t + c 2 e 4t. Since L {5e 4t } = 5 the denominator + 4 i v = + 4 and qv = ( + 2)( + 4) 2. Hence, B qv \ B q = { e 2t, e 4t, te 4t} \ { e 2t, e 4t} = { te 4t}. Math 2065 Section 2 October 12,

4 Therefore, the tet function for y p i y p = Ate 4t. Compute the derivative: y p = A(1 4t)e 4t y p = A( t)e 4t. Subtituting into the differential equation give y p + y p 2y p = A( t)e 4t + 6A(1 4t)e 4t + 8Ate 4t = ( 8 + 6)Ae 4t = 5e 4t. Thu, 2A = 5 o A = 5/2 and y p = ( 5/2)te 4t and the general olution i y g = y h + y p = c 1 e 2t + c 2 e 4t 5 2 te 4t. Math 2065 Section 2 October 12,

5 Laplace Tranform Table f(t) F () = L {f(t)} () t n 3. e at 4. t n e at 5. co bt 6. in bt 7. e at co bt 8. e at in bt 1 n! n+1 1 a n! ( a) n b 2 b 2 + b 2 a ( a) 2 + b 2 b ( a) 2 + b 2 Laplace Tranform Principle Linearity L {af(t) + bg(t)} = al {f} + bl {g} Input Derivative Principle L {f (t)} () = L {f(t)} f(0) L {f (t)} () = 2 L {f(t)} f(0) f (0) Firt Tranlation Principle L {e at f(t)} = F ( a) Tranform Derivative Principle L { tf(t)} () = d d F () The Dilation Principle L {f(bt)} () = 1 L {f(t)} (/b) b Math 2065 Section 2 October 12,

6 Partial Fraction Expanion Theorem The following two theorem are the main partial fraction expanion theorem, a preented in the text. Theorem 1 (Linear Cae). Suppoe a proper rational function can be written in the form p 0 () ( λ) n q() and q(λ) 0. Then there i a unique number A 1 and a unique polynomial p 1 () uch that p 0 () ( λ) n q() = A 1 ( λ) n + p 1 () ( λ) n 1 q(). (1) The number A 1 and the polynomial p 1 () are given by A 1 = p 0(λ) q(λ) and p 1 () = p 0() A 1 q(). (2) λ Theorem 2 (Irreducible Quadratic Cae). Suppoe a real proper rational function can be written in the form p 0 () ( 2 + c + d) n q(), where 2 + c + d i an irreducible quadratic that i factored completely out of q(). Then there i a unique linear term B 1 + C 1 and a unique polynomial p 1 () uch that p 0 () ( 2 + c + d) n q() = B 1 + C 1 ( 2 + c + d) n + p 1 () ( + c + d) n 1 q(). (3) If a + ib i a complex root of 2 + c + d then B 1 + C 1 and the polynomial p 1 () are given by B 1 (a + ib) + C 1 = p 0(a + ib) q(a + ib) and p 1 () = p 0() (B 1 + C 1 )q(). (4) 2 + c + d Math 2065 Section 2 October 12,