Name: Solutions Exam 3

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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 ha been appended to the exam. The following trigonometric identitie may alo be of ue: in(θ + φ) in θ co φ + in φ co θ co(θ + φ) co θ co φ in θ in φ. [0 Point] Find the olution of the following initial value problem: t 2 y + 2ty 2y 0, y() 0, y (). Solution. Thi i a Cauchy-Euler equation with indicial polynomial q() ( ) ( + 4)( 3). Thi q() ha root 4 and 3 o the general olution i y(t) c t 4 + c 2 t 3. Plugging in the initial condition give 0 y() c + c 2 y () 4c + 3c 2. Solving for c and c 2 give c /7 and c 2 /7. Thu, y(t) 7 t t3. 2. [5 Point] Ue variation of parameter to find a particular olution of the nonhomogeneou differential equation y + 4y + 4y t 5 e 2t. You may aume that the olution of the homogeneou equation y + 4y + 4y 0 i y h c e 2t + c 2 te 2t. Solution. Letting y e 2t and y 2 te 2t, a particular olution ha the form y p u y + u 2 y 2 u e 2t + u 2 te 2t, where u and u 2 are unknown function whoe derivative atify the imultaneou equation u e 2t + u 2te 2t 0 2u e 2t + u 2(e 2t 2te 2t ) t 5 e 2t. Math 2065 Section April 30, 200

2 Multiplying both equation by e 2t give u + u 2t 0 2u + u 2( 2t) t 5. Eliminating u give u 2 t 5 and then u tu 2 t 4. Integrating give u (/3)t 3 and u 2 ( /4)t 4 o that y p 3 t 3 e 2t 4 t 4 te 2t. Hence y p 2 t 3 e 2t. 3. [5 Point] Find the Laplace tranform of the following function: f(t) { co t 2 if 0 t < π, if t π. Solution. Ue characteritic function to write f(t) in term of unit tep function: f(t) (co t)χ [0, π) (t) 2χ [π, ) (t) (co t)(h(t) h(t π)) 2h(t π) co t (co t + 2)h(t π). Now apply the econd tranlation theorem to get F () L {f(t)} 2 + e π L {co(t + π) + 2} eπ e π. The + ign in the econd term reult from the fact that co(t + π) co t co(π) in t in π co t. 4. [20 Point] Find the invere Laplace tranform of the following function: (a) F () ( 2) e 3 + ( 2) 2 e 4 Solution. f(t) e 2(t 3) h(t 3) + (t 4)e 2(t 4) h(t 4) Math 2065 Section April 30, 200 2

3 (b) G() e π 2 Solution. g(t) (co 2(t π/2))h(t π/2) (co 2t)h(t π/2) 5. [5 Point] Solve the following initial value problem: y + 4y 6h(t π), y(0) 0, y (0). Solution. Let Y () L {y(t)} where y(t) i the olution of the differential equation and apply the Laplace tranform to both ide of the equation to get Solving for Y () give Ue partial fraction to write which then give Taking invere Laplace tranform give [ ] 5 6. [25 Point] Let A. 2 (a) Compute (I A). Solution. I A Hence, 2 Y () + 4Y () 6 e π. Y () ( 2 + 4) e π. 6 ( 2 + 4) , Y () ( ) e π y(t) in 2t + (4 4 co 2(t π))h(t π) 2 in 2t + (4 4 co 2t)h(t π) { 2 in 2t, if 0 t < π; 2 in 2t co 2t, if t π. 2 [ ] 5 o det(i A) ( )(+) / (I A) [ ] Math 2065 Section April 30, 200 3

4 (b) Find L {(I A) }. Solution. L { (I A) } [ co 3t + in 3t ] 5 in 3t in 3t co 3t. in 3t 3 3 (c) Find the general olution of the ytem y Ay. [ Solution. ] The general olution i L {(I A) } y(0). Thu, if y(0) c, then the general olution i c 2 [ c co 3t + ( y(t) c 3 + 5c ] 3 2) in 3t c 2 co 3t ( 2 c 3 + c ) 3 2 in 3t (d) Solve the initial value problem y Ay, y(0) [ ] 2. Solution. Setting c 2, c 2 in part (c) give [ ] 2 co 3t in 3t y(t). co 3t in 3t Math 2065 Section April 30, 200 4

5 Laplace Tranform Table f(t) F () L {f(t)} (). 2. 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 9. h(t c) n! n+ a n! ( a) n+ 2 + b 2 b 2 + b 2 a ( a) 2 + b 2 b ( a) 2 + b 2 e c 0. δ c (t) e c 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 F () d The Dilation Principle L {f(bt)} () L {f(t)} (/b). b Second Tranlation Principle L {h(t c)f(t c)} e c F (). Math 2065 Section April 30, 200 5

6 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) + p () n ( λ) n q(). () The number A and the polynomial p () are given by A p 0(λ) q(λ) and p () p 0() A 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 + C and a unique polynomial p () uch that p 0 () ( 2 + c + d) n q() B + C ( 2 + c + d) + p () n ( + c + d) n q(). (3) If a + ib i a complex root of 2 + c + d then B + C and the polynomial p () are given by B (a + ib) + C p 0(a + ib) q(a + ib) and p () p 0() (B + C )q(). (4) 2 + c + d Math 2065 Section April 30, 200 6