Name: Solution Exam 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, a table of convolution product, and the tatement of the main partial fraction decompoition theorem have been appended to the exam.. [8 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() in the table. (a) f (t) = e t co 5t Solution. From Formula 0, F () = (b) f (t) = t co 5t Solution. Uing Formula 3 and 8: F () = d L { co 5t} d = d ( ) d + 5 ( ) ( + 5) () = ( + 5) ( ) + 5. = 50 ( + 5). (c) f 3 (t) = e t co 5t (Recall that denote the convolution product.) Solution. Uing Formula 4, 6, and 8: F 3 () = + 5.. [ Point] Compute the invere Laplace tranform of each of the following rational function. 3 + (a) F () = 4 Solution. where F () = A = 3 + + 4 3 + 4 = 3 + ( 6)( + 4) = A 6 + B + 4 = 0 =6 0 = and B = 3 + 6 = 0 = 4 0 =. Hence, (Formula 6) F () = 6 + + 4 o f(t) = e6t + e 4t. Math 065 Section 3 March 3, 008
Name: Solution Exam (b) G() = Solution. 7 + 3 G() = 7 + 3 = 7 ( ) + = ( ) + 6 ( ) +. Hence, (Formula 0 and ) g(t) = e t co t 6 e t in t. 3. [4 Point] Find the characteritic polynomial and the general olution of each of the following contant coefficient linear homogeneou differential equation: (a) y + 8y + y = 0 Solution. The characteritic polynomial i q() = +8+ = (+6)(+), which ha root 6 and. Thu, the general olution of the differential equation i y = c e 6t + c e t. (b) y + 8y + 6y = 0 Solution. The characteritic polynomial i q() = + 8 + 6 = ( + 4), which ha a ingle root 4 of multiplicity. Thu, the general olution of the differential equation i y = c e 4t + c te 4t. (c) y + 8y + 0y = 0 Solution. The characteritic polynomial i q() = + 8 + 0 = ( + 4) + 4, which ha root 4 ± i. Thu, the general olution of the differential equation i y = c e 4t co t + c e 4t in t. (d) y + 4y = 0 Solution. The characteritic polynomial i q() = 3 + 4 = ( + 4), which ha root 0 and ±i. Thu, the general olution of the differential equation i y = c + c co t + c 3 in t. 4. [8 Point Each] Ue the Laplace tranform method to find the olution of the following initial value problem: (a) y 4y + 3y = e t, y(0) = 0, y (0) = 4. Math 065 Section 3 March 3, 008
Name: Solution Exam Solution. Let y(t) be the unknown olution function and let Y () = L {y(t)} be the Laplace tranform of y(t). Taking the Laplace tranform of both ide of the differential equation and uing Formula, 3, and 6, we get Y 4 4Y + 3Y =. Rewrite thi a ( 4 + 3)Y = 4 + = 4 3, and olve for Y to get 4 3 Y = ( 3)( ). To compute y(t) = L {Y ()}, expand Y in a partial fraction uing the ( )- chain: 4 3 Y = ( 3)( ) = A ( ) + p () ( )( 3), where A = 4 3 3 = = and p () = (4 3) + ( 3) Continuing the partial fraction calculation give: Y = 9 ( ) + ( )( 3) = 9 9 ( ) 4 + 4 3. = 9 9 = 9. Then y(t) = L {Y ()} = tet 9 4 et + 9 4 e3t. (b) y + y = co t, y(0) = 0, y (0) = 0. Solution. Let y(t) be the unknown olution function and let Y () = L {y(t)} be the Laplace tranform of y(t). Taking the Laplace tranform of both ide of the differential equation and uing Formula 3 and 8, we get Solve for Y to get Y = Y + Y = Then ue Formula 8, 9, and 4 to get +. ( + ) = + y(t) = L {Y ()} = L { + +. } = in t co t = t in t, + where the lat equality come from the convolution table (3rd formula from the bottom). Math 065 Section 3 March 3, 008 3
Name: Solution Exam Exam II Supplementary Sheet A Short Table of Laplace Tranform. L {af(t) + bg(t)} () = af () + bg(). L {e at f(t)} () = F ( a) 3. L { tf(t)} () = d d F () 4. L {} () = 5. L {t n } () = n! n+ 6. L {e at } () = 7. L {t n e αt } () = 8. L {co bt} () = 9. L {in bt} () = 0. L {e at co bt} () =. L {e at in bt} () = a n! ( α) n+ + b b + b a ( a) + b b ( a) + b. L {f (t)} () = F () f(0) 3. L {f (t)} () = F () f(0) f (0) 4. L {(f g)(t)} () = F ()G() Math 065 Section 3 March 3, 008 4
Name: Solution Exam Table of Convolution f(t) g(t) f g(t) t t n t n+ t t t (n + )(n + ) at a a ( ( a t )) 3 t (at ) a3 t e at e at ( + at) a t e at a 3 (eat (a + at + a t )) e at e bt b a (ebt e at ) e at e at te at e at e at in bt co bt in bt co bt co bt a a b a + b (beat b co bt a in bt) a + b (aeat a co bt + b in bt) (b a in bt) b a a b ( at ) a (a a co bt) b a a b t (a b in bt) a b a b (at + ) a Math 065 Section 3 March 3, 008 5
Name: Solution Exam Partial Fraction Expanion Theorem The following two theorem are the main partial fraction expanion theorem, a preented in the text. Theorem (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 and a unique polynomial p () uch that p 0 () ( λ) n q() = A ( λ) n + p () ( λ) n q(). () The number A and the polynomial p () are given by A = p 0() q() and p () = p 0() A q(). () =λ λ Theorem (Irreducible Quadratic Cae). Suppoe a real proper rational function can be written in the form p 0 () ( + c + d) n q(), where + c + d i an irreducible quadratic that i factored completely out of q(). Then there i a unique linear term B + C and a unique polynomial p () uch that p 0 () ( + c + d) n q() = B + C ( + c + d) n + p () ( + c + d) n q(). (3) If a + ib i a complex root of + c + d then B + C and the polynomial p () are given by B + C =a+bi = p 0() q() and p () = p 0() (B + C )q(). (4) =a+bi + c + d Math 065 Section 3 March 3, 008 6