SOLUTIONS FOR HOMEWORK SECTION 6.4 AND 6.5

Transcription

1 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND 6.5 Problem : For each of the following function do the following: (i) Write the function a a piecewie function and ketch it graph, (ii) Write the function a a combination of term of the form u a (t)k(t a) and compute the Laplace tranform (a) f(t) = t( u (t)) + e t (u (t) u 2 (t)) (b) h(t) = in(2t) + u π (t)(t/π in(2t)) + u 2π (t)(2π t)/π (c) g(t) = u 0 (t) + 5 k= ( )k u k (t) (a) The graph i ketched in figure. t 0 t < f(t) = e t t < 2 0 t 2 Figure. graph of f(t) To find the Laplace tranform of f(t), rewrite f(t) a f(t) = t + (e t t)u (t) e t u 2 (t) L{f} = L{t} + L{e t u (t)} L{tu (t)} L{e t u 2 (t)}

2 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND Apply t-hifting theorem (Theorem 6.3. in the textbook) or -hifting theorem if needed, finding the Laplace tranform for each term: L{t} = 2 L{e t u (t)} = el{e (t ) u (t)} = e e L{e t } = e( ) L{tu (t)} = L{(t )u (t)} + L{u (t)} = e 2 + e L{e t u 2 (t)} = e 2 L{e (t 2) u 2 (t)} = e 2 e 2 L{e t } = e(2 2) Combining all term yield L{f} = + e( ) 2 e e 2 e(2 2) (b) in(2t) 0 t π h(t) = t/π π t < 2π 2 π 2 The graph i ketched in figure 2. Figure 2. graph of h(t) To find the Laplace tranform of h(t), L{h} = L{in(2t)} + L{u π (t)(t/π in(2t))} + L{u 2π (t)(2π t)/π} Apply t-hifting theorem if needed, and find the Laplace tranform of each term: L{in(2t)} =

3 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND L{u π (t)(t/π in(2t))} = e π L{( t + π in(2(t + π)))} = e π ( π π ) L{u 2π (t)(2π t)/π} = e 2π 2π (t + 2π) L{ } = e 2π π π 2 Combining all term yield L{h} = e π ( π ) e 2π π 2 (c) Here u 0 (t) = ince we are only intereted in the domain t 0. 0 t < 0 t < 2 2 t < 3 g(t) = 0 3 t < 4 4 t < 5 0 t 5 The graph i ketched in figure 3. Figure 3. graph of g(t) To find the Laplace tranform of g(t), L{g} = L{} + = + 5 k= = e 5 L{( ) k u k (t)} k= ( ) k e k + e 2 e 3 + e 4 e 5 Problem 2: Solve the initial value problem 0 if 0 t < y + 6y = g(t) where g(t) = 2 if t < 7 0 if 0 t

4 with initial value y(0) = 4 Ue tep function to repreent g(t) a SOLUTIONS FOR HOMEWORK SECTION 6.4 AND g(t) = 2(u (t) u 7 (t)) Take the Laplace tranform of the differential equation and plug in initial value to get Y () 4 + 6Y () = 2( e e 7 ) Solving for Y () yield Y () = 2e ( + 6) 2e 7 ( + 6) Let H() =, by partial fraction, we can rewrite H() a (+6) H() = 6 ( + 6 ) The invere Laplace tranform of H() i h(t) = L {H} = 6 ( e 6t ) Hence, by uing t-hifting theorem if neceary, we can find the olution y(t) = L{Y } = 2u (t)h(t ) 2u 7 (t)h(t 7) + 4e 6t Problem 3: Solve the initial value problem and with y(0) = 0 and y (0) = 0. y + y = g(t) where g(t) = { t if 0 t < 0 if t Firtly we can rewrite g(t) a g(t) = t( u (t)) To find the Laplace tranform of g(t), we need ue t-hifting theorem L{g(t)} = L{t} L{u (t)t} = e L{t + } = e ( 2 + ) Then take the Laplace tranform of the differential equation and plug in initial value to get 2 Y () + Y () = e e 2 Solving for Y () yield Y () = ( 2 + ) e ( 2 + ) e 2 ( 2 + )

5 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND Let H() = and F () = ( 2 + ) 2 ( 2 + ) We could ue the method of undetermined coefficient method to find the partial fraction for H(): H() = and F () = The invere Laplace tranform of them are h(t) = co(t) and f(t) = t in(t) By applying t-hifting theorem if neceary, the invere Laplace tranform of Y () i: y(t) = L{H()} L{e H()} L{e F ()} = h(t) u (t)h(t ) u (t)f(t ) Problem 4: Conider the ma-pring ytem decribed by the initial value problem y + 4y = in t + u π/2 (t) co t; y(0) = 0, y (0) = 0. Find the olution of the initial value problem. Hint: Rewrite co t uing a trigonometric identity. Firt need to write u π/2 (t) co t in the form u c (t)f(t c). To do thi ue the trig identitie co θ = in(π/2 θ) and in( θ) = in θ. Thi give ( π ) ( u π/2 (t) co t = u π/2 (t) in 2 t = u π/2 (t) in t π ) 2 Subtituting thi into the ODE and taking the Laplace tranform give u Solving for Y yield Let ( 2 + 4)Y () = 2 + e π Y () = π ( ( 2 + )( 2 e 2 ) + 4) H() = ( 2 + )( 2 + 4) We ue partial fraction to rewrite H: ( 2 + )( 2 + 4) = A + B C + D 2 + = = (A + B)( 2 + ) + (C + D)( 2 + 4) = = (A + C) 3 + (B + D) 2 + (A + 4C) + (B + 4D) Equating coefficient yield the four equation A + C = 0, B + D = 0, A + 4C = 0, B + 4D =

6 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND which ha the olution A = C = 0, B = /3 and D = /3. Therefore we can write and o H() = h(t) = L {H} = 6 in(2t) + 3 in(t). It follow (uing the t-hifting theorem where neceary) that y(t) = h(t) u π/2 (t)h(t π/2) Problem 5: Determine the value of the following integral (a) 7 δ(t + ) dt. 2 (b) 7 δ(t ) dt. 2 (c) 8 ln (t2 )δ(t e) dt. (d) 4 2 (2t4 t 3 + 7t 2 ) δ(t 5) dt. To evaluate the above integral, we will be uing the well-known formula derived in cla which i given by δ(t t 0 )f(t) dt = f(t 0 ). It hould be noted in paing that the value of t 0 mut lie within the interval of integration. Otherwie, the value of the definite integral will be zero. (a) 0 (b) (c) 2 (d) 0 Problem 6: Find the olution of the initial value problem and ketch the graph of the olution for y + 2y + 2y = δ(t π); y(0) =, y (0) = 0. (0.)

7 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND Let u take initially the Laplace tranform of the IVP given and olve for Y () = L{y(t)} afterward: 2 Y + 2Y 2 + 2Y = e π + 2 Y = e π Y = ( + ) e π ( + ) 2 + Y = + ( + ) ( + ) e π ( + ) 2 +. (0.2) Next, note that the invere of the firt two term in Eq. (0.2) can be obtained by utilizing the -hifting theorem, that i, { } + { } L ( + ) 2 = e t L = e t co (t), (0.3) { } { } L ( + ) 2 = e t L = e t in (t). (0.4) Note that we could ue the formula 9 and 0 in p.g. 252 in the textbook for the above invere. Finally, and a far a the lat term in Eq. (0.2) i concerned, we will apply a combination of the t- and - hifting theorem: { } } = L{H( + )e π L e π ( + ) 2 + = u π (t)h(t π), (0.5) where h(t) = e t in (t) and u π repreent the Heaviide function for c = π. A a ide note, H() = 2 + and h(t) = L{H( + )} = e t in (t), ee Eq. (0.4). Thu, upon combining all the above term, the olution to the IVP i given by y = y(t) = e t [co (t) + in (t)] u π (t)e (t π) in (t). (0.6) The graph of the olution given by Eq. (0.6) i hown in Fig Figure 4. Graph of the olution to the IVP of Quetion.

8 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND Problem 7: Find the olution of the initial value problem and ketch the graph of the olution for y + y = δ(t 2π)co(t); y(0) = 0, y (0) = Take the Laplace tranform of the equation to get 2 Y () y(0) y (0) + Y () = L{δ(t 2π)co(t)} To find L{δ(t 2π)co(t)}, by definition of the Laplace tranform and ue the property of delta function to get L{δ(t 2π)co(t)} = Plugging initial value yield Solving for Y() to get 0 δ(t 2π)co(t)e t dt = co(2π)e 2π = e 2π ( 2 + )Y () = e 2π Y () = e 2π Let H() =, then h(t) = 2 + L {H} = in(t). Apply t-hifting theorem to get So the olution i L {e 2π H()} = u 2π (t)h(t 2π) = u 2π (t)in(t 2π) = u 2π (t)in(t) The graph i ketched in figure 5 y(t) = L {Y } = u 2π (t)in(t) + in(t) Figure 5. Graph of the olution Problem 8: Conider the initial value problem y + 2y + y = kδ(t ), y(0) = 0, y (0) = 0. (a) Find the olution of the initial value problem. (b) Find the value of k for which the repone ha a peak value of 2.

9 SOLUTIONS FOR HOMEWORK SECTION 6.4 AND (a) Take the Laplace tranform of the differential equation to get ( )Y () = ke o that after noting that = ( + ) 2 and olving for Y we have Y () = ke 2 ( + ) 2. We recognize /(+) 2 a / 2 hifted by the value c =. Applying the -hifting theorem give u { } L = te t ( + ) 2 Now applying the t-hifting theorem give u the olution y(t) = k u (t) (t )e t+ (b) A y(t) = 0 for t <, we ignore thi region and look for extreme value on the interval t. For thi time range, we may replace u (t) with. Now and etting equal to zero give the equation y (t) = ke t+ k(t )e t+ ke t+ k(t )e t+ = 0 = ke t+ (2 t) = 0 o that we conclude the extreme value occur at t = 2. Computing y(2) and etting equal to 2 give u the equation for k: k = 2 = k = 2e e