Name: Solutions Exam 3

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1 Instructions. Answer each of the questions on your own paper. Put your name on each page of your paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. A table of Laplace transforms has been appended to the exam. The following trigonometric identities may also be of use: sin(θ+ϕ) = sinθcosϕ+sinϕcosθ cos(θ+ϕ) = cosθcosϕ sinθsinϕ. [20 Points] Find the general solution of the following Cauchy-Euler equations: (a) 3t 2 y 7ty +3y = 0. Solution. The indicial polynomial is Q(s) = 3s(s ) 7s+3 = 3s 2 0s+3 = (3s )(s 3), which has the two distinct real roots /3 and 3. Hence the general solution is (b) t 2 y ty +0y = 0. Solution. The indicial polynomial is y = c t /3 +c 2 t 3. Q(s) = s(s ) s+ = s 2 2s+0 = (s ) 2 +9, which has the complex roots ±3i. Hence the general solution is y = c tcos(3ln t )+c 2 tsin(3ln t ). 2. [20 Points] Use variation of parameters to find a particular solution of the nonhomogeneous differential equation y 4y +4y = t /2 e 2t. You may assume that the solution of the homogeneous equation y 4y +4y = 0 is y h = c e 2t +c 2 te 2t. Solution. Letting y = e 2t and y 2 = te 2t, a particular solution has the form y p = u y +u 2 y 2 = u e 2t +u 2 te 2t, Math 2065 Section 2 November 2, 208

2 where u and u 2 are unknown functions whose derivatives satisfy the simultaneous equations u e2t +u 2 te2t = 0 u (2e 2t +u 2(e 2t +2te 2t ) = t /2 e 2t. Dividing both equations by e 2t gives the simpler equations u +u 2 t = 0 2u +u 2 (+2t) = t/2. Applying Cramer s rule gives 0 t u = t /2 +2t t 2 +2t and 0 u 2 t /2 2 = t 2 +2t = t3/2 = t 3/2 = t/2 = t/2 Integrating gives u = 2 5 t5/2 and u 2 = 2 3 t3/2 so that y p = 2 5 t5/2 e 2t t3/2 te 2t = 4 5 t5/2 e 2t. 3. [20 Points] Let f be the function defined by t 2 if 0 t < 2, f(t) = 3 if t 2. (a) Sketch the graph of f(t) over the interval [0, 4] Math 2065 Section 2 November 2, 208 2

3 (b) Find the Laplace transform of f(t). Solution. Use characteristic functions to write f(t) in terms of unit step functions: f(t) = t 2 χ [0,2) (t)+3χ [2, ) (t) = t 2 (h(t) h(t 2))+3h(t 2) = t 2 +(3 t 2 )h(t 2). Now apply the second translation theorem to get F(s) = Lf(t)} = 2 s 3 +e 2s L 3 (t+2) 2} = 2 s 3 +e 2s L 3 (t 2 +4t+4) } = 2 s 3 +e 2s L t 2 4t ) } = 2 s 3 e 2s ( 2 s s 2 + s 4. [20 Points] Find the inverse Laplace transform of the following functions: (a) F(s) = e 4s (s+2) 3 Solution. F(s) = F (s)e 4s where F (s) = ). (s+2) 3. Let f (t) = L F (s)} = 2 t2 e 2t. Then the inverse of the second translation theorem gives f(t) = L F(s)} = f (t 4)h(t 4) (b) G(s) = 2s s 2 +2s+5 e 2s = 2 (t 4)2 e 2(t 4) h(t 4) 0 if 0 t < 4 = 2 (t 4)2 e 2(t 4) if t 4. Math 2065 Section 2 November 2, 208 3

4 Solution. Let G (s) = G (s) = 2s s 2 +2s+5. Then 2s s 2 +2s+5 = 2s (s+) 2 +4 = 2(s+) 2 (s+) 2 +4 = 2(s+) (s+) (s+) Thus, taking the inverse Laplace transform gives g (t) = L G (s)} = 2e t cos2t e t sin2t, so that the inverse of the second translation theorem gives g(t) = L G(s)} = g (t 2)h(t 2) ( ) = 2e (t 2) cos2(t 2) e (t 2) sin2(t 2) h(t 2) 0 if 0 t < 2 = 2e (t 2) cos2(t 2) e (t 2) sin2(t 2) if t [20 Points] Solve the following initial value problem: y +y = δ(t π), y(0) = 0, y (0) =. (Remember that δ(t c) is the Dirac delta function centered at c.) Give a careful sketch of the graph of the solution for the interval 0 t 2π. Solution. Let Y(s) = Ly(t)} be the Laplace transform of the solution function. Apply the Laplace transform to both sides of the equation to get Thus, s 2 Y(s) +Y(s) = e πs. (s 2 +)Y(s) = +e πs, and hence Y(s) = s 2 + +e πs s 2 +. Apply the inverse Laplace transform to get y(t) = sint+h(t π)sin(t π) = sint (sint)h(t π) sint if 0 t < π, = 0 if t π. Math 2065 Section 2 November 2, 208 4

5 π 2π 3π Graph of y(t). Math 2065 Section 2 November 2, 208 5

6 Laplace Transform Table f(t) F(s) = Lf(t)}(s). 2. t n 3. e at 4. t n e at 5. cos bt 6. sin bt 7. e at cosbt 8. e at sinbt 9. h(t c) s n! s n+ s a n! (s a) n+ s s 2 +b 2 b s 2 +b 2 s a (s a) 2 +b 2 b (s a) 2 +b 2 e sc 0. δ c (t) = δ(t c) e sc s Laplace Transform Principles Linearity Input Derivative Principles First Translation Principle Transform Derivative Principle Laf(t)+bg(t)} = alf}+blg} Lf (t)}(s) = slf(t)} f(0) Lf (t)}(s) = s 2 Lf(t)} sf(0) f (0) Le at f(t)} = F(s a) L tf(t)}(s) = d ds F(s) Second Translation Principle Lh(t c)f(t c)} = e sc F(s), or Lg(t)h(t c)} = e sc Lg(t+c)}. Math 2065 Section 2 November 2, 208 6

7 Partial Fraction Expansion Theorems The following two theorems are the main partial fractions expansion theorems, as presented in the text. Theorem (Linear Case). Suppose a proper rational function can be written in the form (s λ) n q(s) and q(λ) 0. Then there is a unique number A and a unique polynomial p (s) such that (s λ) n q(s) = A (s λ) + p (s) n (s λ) n q(s). () The number A and the polynomial p (s) are given by A = p 0(λ) q(λ) and p (s) = p 0(s) A q(s). (2) s λ Theorem 2 (Irreducible Quadratic Case). Suppose a real proper rational function can be written in the form (s 2 +cs+d) n q(s), where s 2 + cs + d is an irreducible quadratic that is factored completely out of q(s). Then there is a unique linear term B s+c and a unique polynomial p (s) such that (s 2 +cs+d) n q(s) = B s+c (s 2 +cs+d) n + p (s) (s s +cs+d) n q(s). (3) If a+ib is a complex root of s 2 +cs+d then B s+c and the polynomial p (s) are given by B (a+ib)+c = p 0(a+ib) q(a+ib) and p (s) = p 0(s) (B s+c )q(s). (4) s 2 +cs+d Math 2065 Section 2 November 2, 208 7