Questions and Answers #7 Laplace Transform

Questions and Answers Sheet 7 Laplace Transform Question #1 Using the derivative theorem and, find the Laplace transform of the following functions 𝑟 cos ℎ (𝑡) Answer: Laplase transform of 𝑡 cos(ℎ) 𝑡 𝐿(𝑡 cos(ℎ) 𝑡) derivative of Laplase transform of cos(ℎ) 𝑡 𝑠 ∴ 𝐿(cos ℎ 𝑡) = 2 𝑠 −1 𝑑 𝑠 ∴ 𝐿(𝑡 cos ℎ 𝑡) = −𝑠 2 −1 = (𝑠2 −1)2 𝑑𝑠 𝑠 2 −1 = 𝑠 2 −1−2𝑠 2 (𝑠 2 −1)2 (−1)2 (−1) 𝑠 2 +1 = (𝑠2 −1)2 Question #2 Find the Laplace transform of 4𝑡 2 + 5𝑡 3 Answer: We know that 𝑛! 𝐿[𝑡 𝑛 ] = 𝑛+1 𝑠 𝐿[4𝑡 2 + 5𝑡 3 ] = 𝐿[4𝑡 2 ] + 𝐿[5𝑡 3 ] = 4𝐿[𝑡 2 ] + 5𝐿[𝑡 3 ] =4⋅ = 8 𝑠3 2! +5⋅ 𝑠3 30 + 3! 𝑠4 𝑠4 Question #3 Consider the IVP y''-y'=0 where y(0)=2 and y'(0)=3.Solve it using Laplace transforms Answer: Step 1 𝑦 ′′ − 𝑦 ′ = 0 ⋯ (1) where y(0)=2 and y'(0)=3 Step 2 Using Laplace transform 𝐿{𝑦 ′ (𝑡)} = 𝑠𝐿{𝑦(𝑡)} − 𝑦(0) And 𝐿{𝑦 ′′ (𝑡)} = 𝑠 2 𝐿{𝑦(𝑡)} − 𝑠𝑦(0) − 𝑦 ′ (0) Step 3 Applying the Laplace transform on both sides in (1) 𝐿{𝑦 ′′ (𝑡)} − 𝐿{𝑦 ′ (𝑡)} = 𝐿{0} 𝑠 2 𝐿{𝑦(𝑡)} − 𝑠𝑦(0) − 𝑦 ′ (0) − 𝑠𝐿{𝑦(𝑡)} + 𝑦(0) = 0 𝑠 2 𝐿{𝑦(𝑡)} − 2𝑠 − 3 − 𝑠𝐿{𝑦(𝑡)} + 2 = 0 𝑠 2 𝐿{𝑦(𝑡)} − 2𝑠 − 𝑠𝐿{𝑦(𝑡)} − 1 = 0 𝐿{𝑦(𝑡)}(𝑠 2 − 𝑠) = 1 + 2𝑠 1+2𝑠 𝐿{𝑦(𝑡)} = ⋯ (2) 𝑠(𝑠−1) Step 4 The partial fraction decomposition of 1+2𝑠 𝐴 𝐵 = + 𝑠(𝑠−1) 𝑠 𝑠−1 1 + 2𝑠 = 𝐴(𝑠 − 1) + 𝐵𝑠 1 + 2𝑠 = 𝐴𝑠 − 𝐴 + 𝐵𝑠 Equating the coefficient of s and constant term A+B=2 and A=-1 ⇒ −1 + 𝐵 = 2 ⇒𝐵=3 1 + 2𝑠 1 3 =− + 𝑠(𝑠 − 1) 𝑠 𝑠−1 Step 5 From (2) 1 𝐿{𝑦(𝑡)} = − + 3 𝑠 −1 𝑠−1 1 3 ⇒ 𝑦(𝑡) = 𝐿 {− + } 𝑠 𝑠−1 1 3 ⇒ 𝑦(𝑡) = 𝐿−1 {− } + 𝐿−1 { } 𝑠 𝑠−1 1 1 ⇒ 𝑦(𝑡) = −𝐿−1 { } + 3𝐿−1 { } 𝑠 𝑠−1 ⇒ 𝑦(𝑡) = −𝑡 + 3𝑒 𝑡 Question #4 Find the inverse Laplace Transform for the following expressions 5𝑠 (𝑠 − 2)2 Answer: And: 5𝑒 2𝑡 (2𝑡 + 1) We use following formula 𝐿−1 ( 1 ) = 𝑒 𝑎𝑡 𝑠−𝑎 1 𝐿−1 ((𝑠−𝑎)𝑛 ) = 5𝑠 (𝑠−2)2 5 = = 𝑒 𝑎𝑡 𝑡 𝑛−1 (𝑛−1)! 5(𝑠−2)+10 + (𝑠−2)2 10 (𝑠−2)2 𝑠−2 5𝑠 5 10 −1 𝐿 {(𝑠−2)2} = 𝐿−1 { + (𝑠−2)2} 𝑠−2 1 1 −1 −1 = 5𝐿 ( 𝑠−2 ) + 10𝐿 = 5𝑒 2𝑡 + 10 ((𝑠−2)2 ) 𝑒 2𝑡 𝑡 2−1 (2−1)! = 5𝑒 2𝑡 + 10𝑡𝑒 2𝑡 = 5𝑒 2𝑡 (1 + 2𝑡) 5𝑠 𝐿−1 {(𝑠−2)2} = 5𝑒 2𝑡 (1 + 2𝑡) Question #5 Laplace transform u(t) is the unit step function Find the Laplace transform of the following continuous-time signal. 𝑥(𝑡) = 4𝑡𝑒 −5𝑡 𝑢(𝑡) Answer: Step 1 The given signal is 𝑥(𝑡) = 4𝑡𝑒 −5𝑡 𝑢(𝑡) Step 2 The above signal can be compared to 𝑥(𝑡) = 𝑓(𝑡 − 𝑐)𝑢(𝑡 − 𝑐) with c=0 and 𝑓(𝑡 − 𝑐) = 4𝑡𝑒 −5𝑡 but, the laplace transform 𝐿{𝑓(𝑡 − 𝑐)𝑢(𝑡 − 𝑐)} = 𝑒 −𝑎𝑠 𝐹(𝑠) where F(s) is the laplace transform of f(t). Here, 𝑓(𝑡 − 0) = 𝑓(𝑡) = 4𝑡𝑒 −5𝑡 Hence, 𝐿{𝑥(𝑡)} = 𝐿{𝑓(𝑡)𝑢(𝑡)} 𝐿{𝑥(𝑡)} = 𝑒 0 𝐿{𝑓(𝑡)} 𝐿{𝑥(𝑡)} = 𝐿{4𝑡𝑒 −5𝑡 } 𝐿{𝑥(𝑡)} = 4𝐿{𝑡𝑒 −5𝑡 } 4 𝑛 𝐿{𝑥(𝑡)} = (𝑠+5)2 [∵ 𝐿{𝑡 𝑛 𝑒 𝑎𝑡 } = (𝑠−𝑎)𝑛+1] 4 Hence , the laplace transform of the given signal is 𝑋(𝑠) = (𝑠+5)2 Question #6 Find Laplace transforms of sin ℎ 3𝑡 cos 2 2 𝑡 Answer: 𝑓(𝑡) = cos 2 2 𝑡 applying laplace transform (1+cos 4𝑡) 𝐿{cos 2 2 𝑡} = 𝐿{ 1 } 2 1 = 𝐿{1} + 𝐿{cos 4 𝑡} 2 1 1 2 1 = [ ]+ [ 2 𝑠 𝑠 2 𝑠 2 +4 2 1 𝐿{cos 2 2 𝑡} = 2𝑠 ] + 𝑠 2(𝑠 2 +16) 2) 𝑓(𝑡) = sin ℎ 3𝑡 𝐿{𝑓(𝑡)} = 𝐿{sin ℎ 3𝑡} = 3 𝑠 2 −32 = 3 𝑠 2 −9 Question #7 Find the solution of the integral equation f(t) using Laplace transforms 𝑡 𝑓(𝑡) = 36𝑡 + 5 ∫ 𝑓(𝑡 − 𝑢) sin(5𝑢) 𝑑𝑢 0 Answer: 𝑡 𝑓(𝑡) = 36𝑡 + 5 ∫0 𝑓(𝑡 − 𝑢) sin(5𝑢) 𝑑𝑢 ⇒ 𝑓(𝑡) = 36𝑡 + 5𝑓(𝑡) ⋅ sin(5𝑡) ⇒ 𝐿{𝑓(𝑡)} = 36𝐿{𝑡} + 5𝐿{𝑓(𝑡) ⋅ sin(5𝑡)} 36 ⇒ 𝐹(𝑠) = 2 + 5𝐿{𝑓(𝑡)}𝐿{sin(5𝑡)} = 36 𝑠2 𝑠 + 5𝐹(𝑠) 5 𝑠 2 +25 ⇒ 𝐹(𝑠) − 5𝐹(𝑠) 5 𝑠 2 +25 = 36 𝑠2 5𝑠 ⇒ 𝐹(𝑠) [1 − 𝑠 2 +25 𝑠 2 −5𝑠+25 ⇒ 𝐹(𝑠) [ ]= ⇒ 𝐹(𝑠) = now, 36 ]= 𝑠2 36 𝑠 2 +25 𝑠2 36 𝑠 2 +25 × 𝑠2 𝑠 2 −5𝑠+25 36(𝑠 2 +25) 𝐴 = + 𝑠 2 (𝑠 2 −5𝑠+25) 2 𝑠 𝐵 𝑠2 + 𝐶𝑠+𝐷 𝑠 2 −5𝑠+25 ⇒ 36(𝑠 + 25) = 𝐴𝑠(𝑠 2 − 5𝑠 + 25) + 𝐵(𝑠 2 − 5𝑠 + 25) + (𝐶𝑠 + 𝐷)𝑠 2 𝑠 = 0,36(25) = 𝐵(25) ⇒ 𝐵 = 36 coefficient of 𝑠 2 , 36 = 𝐴 − 5𝐴 + 𝐵 + 𝐷 coefficient of 𝑠 3 , 0 = 𝐴 + 𝐶 coefficient of s, 0 = 25𝐴 + (−5)𝐵 36 −36 180 ⇒ 𝐴 = ,𝐶 = and 𝐷 = 5 ∴ 𝐹(𝑠) = 5 36(𝑠 2 +25) 𝑠 2 (𝑠 2 −5𝑠+25) ⇒ 𝐹(𝑠) = ⇒ 𝑓(𝑡) = 36 1 5 𝑠 36 5 + + 𝑠 𝑠 𝑠2 2 + 5√3 2 + 36𝑡 − 𝑡) + 18 ie) 𝑓(𝑡) = cos ( 5√3 2 36 5 𝑡) + 5 2 75 ) 2 4 5((𝑠− ) + 36 5 5 𝑒 2𝑡 𝐿−1 { 5 2 5√3 𝑒 2𝑡 sin ( + 36𝑡 − 12√3 5 −36𝑠+180 5(𝑠 2 −5𝑠+25) −36(𝑠−2)+18(5) 36 36 5𝑡 ⇒ 𝑓(𝑡) = + 36𝑡 − 𝑒2 5 5 cos ( + 5 36 𝑠 5 36 36 1 = 5 𝑡 2 36 5 𝑒 5√3 2 𝑠 75 𝑠2+ 4 }+ 5 18(5) −1 1 𝐿 { 2 75}𝑒 2𝑡 5 𝑠 + 4 𝑡) 5 𝑡 2 𝑒 sin ( 5√3 2 𝑡) Question #8 Solve the differential equation using either the method of undetermined coefficients or Laplace transforms. 𝑦 ′′′ − 6𝑦 ′′ + 11𝑦 ′ − 6𝑦 = 3𝑥 Answer: Step 1 The differential equation is 𝑦 ′′′ − 6𝑦 ′′ + 11𝑦 ′ − 6𝑦 = 3𝑥 Obtain the auxiliary equation as follows. 𝑚3 − 6𝑚2 + 11𝑚 − 6 = 0 (m-1)(m-2)(m-3)=0 m=1,2 and 3 The roots are 𝑚1 = 1, 𝑚2 = 2 and 𝑚3 = 3 Thus , the complimentary solution of the homogeneous equation is 𝑦𝑐 (𝑥) = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥 Step 2 Obtain the particular solution as follows. Let 𝑦𝑝 (𝑥) = 𝐴𝑥 + 𝐵 Substitute 𝑦𝑝 (𝑥) = 𝐴𝑥 + 𝐵 in 𝑦 ′′′ − 6𝑦 ′′ + 11𝑦 − 6 = 3𝑥 (0) − 6(0) + 11(𝐴) − 6(𝐴𝑥 + 𝐵) = 3𝑥 11A-6Ax-6B=3x -6Ax=3x, 11A-6B=0 (By comparing like terms) 3 11 𝐴 = − ,𝐵 = − 6 12 1 11 2 12 Thus, the particular solution is 𝑦𝑝 (𝑥) = − 𝑥 − Therefore , the general solution is 1 11 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥 − 𝑥 − 2 12 Step 3 Answer 1 11 The general solution of the differential equation is 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥 − 𝑥 − 2 Question #9 find the Laplace transform of f (t). 𝑓(𝑡) = 𝑒 𝑡 −𝑒 −𝑡 𝑡 Answer: Step 1 Laplace transform of common function: 𝐿{𝑒 𝑎𝑡 𝑓(𝑡)} = 𝐹(𝑠 − 𝑎) ∞ 𝑓(𝑡) 𝐿{ 𝑡 } = ∫𝑠 𝐹(𝑠)𝑑𝑠 Step 2 𝑒 𝑡 − 𝑒 −𝑡 𝑡 ∞ 𝐹(𝑠) = ∫𝑠 𝐿{𝑒 𝑡 − 𝑒 −𝑡 }𝑑𝑠 𝑓(𝑡) = ∞ = ∫𝑠 ( = 1 − 1 ) 𝑑𝑠 𝑠−1 𝑠+1 ∞ 1 ∞ 1 ∫𝑠 𝑠−1 𝑑𝑠 − ∫𝑠 𝑠+1 𝑑𝑠 𝐹(𝑠) = [ln(𝑠 − 1) − ln(𝑠 + 1)]∞ 𝑠 𝐹(𝑠) = [ln ( 𝑠−1 𝑠+1 ∞ )] 𝑠 𝑠−1 𝐹(𝑠) = ln 1 − ln ( 𝐹(𝑠) = ln ( 𝑠+1 𝑠−1 𝑠+1 ) ) (∴ − ln 𝑎 = ln 1 𝑎 ) Question #10 Use the Laplace Transform to solve the given initial value problem. 𝑦 ′′ + 𝑦 = √2 sin(√2𝑡) , 𝑦(0) = 7, 𝑦 ′ (0) = 2 Answer: Step 1 Laplace transform of derivative of f: The laplace transform of second derivative of f is given by: 𝐿(𝑓 ′′ (𝑡)) = 𝑠 2 𝐹(𝑠) − 𝑓(0) − 𝑓 ′ (0) Given: 𝑦 ′′ + 𝑦 = √2 sin(√2𝑡) Take Laplace transform on both sides 𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) + 𝑦(𝑠) = √2 ( Here, L(y(t))=y(s) and 𝐿(sin 𝑎 𝑡) = √2 𝑠 2 +2 𝑎 ) 𝑠 2 +𝑎2 Plug in the initial conditions y(0)=7 , y'(0)=2 2 𝑠 2 𝑌(𝑠) − 7𝑠 − 2 + 𝑌(𝑠) = 2 (𝑠 2 + 1)𝑌(𝑠) = 2 𝑠 2 +2 𝑠 +2 + 7𝑠 + 2 12 2 𝑌(𝑠) = (𝑠2 𝑌(𝑠) = +2)(𝑠 2 +1) 2 2 𝑠 2 +1 𝑌(𝑠) = 4 ( − 𝑠 2 +2 1 𝑠 2 +1 + 7( 𝑠 𝑠 2 +1 𝑠 + 7( ) − √2 )+ 𝑠 2 +1 √2 𝑠 2 +2 )+ + 7( 2 𝑠 2 +1 2 𝑠 2 +1 𝑠 𝑠 2 +1 ) Take Laplace inverse transform on both sides 𝑦(𝑡) = 4 sin 𝑡 − √2 sin √2 𝑡 + 7𝑐𝑜𝑠 Step 2 Answer: 𝑦(𝑡) = 4 sin 𝑡 − √2 sin √2 𝑡 + 7 𝑐𝑜𝑠 Question #11 Find the inverse Laplace transform of 1 𝐺(𝑠) = 2 𝑠 −4𝑠+5 Answer: Step 1 Calculation Function is given as 1 𝑠 2 −4𝑠+5 to find laplace inverse transform 1 1 = 2 2 𝑠 −4𝑠+5 1 𝑠 2 −4𝑠+5 𝑠 −4𝑠+4+1 1 = (𝑠−2)2 +1 Step 2 Applying inverse laplace transform rule: 𝐿{𝐹(𝑠)} = 𝑓(𝑡) then 𝐿−1 {𝐹(𝑠 − 𝑎)} = 𝑒 𝑎𝑡 𝑓(𝑡) so, 1 1 𝐿−1 {(𝑠−2)2 } = 𝑒 2𝑡 𝐿−1 { 2 } +1 𝑠 +1 1 −1 𝐿 { } = 𝑠𝑖𝑛 1 𝑒 −2𝑡 𝐿−1 { 2 } = 𝑒 2𝑡 𝑠𝑖𝑛 𝑠 +1 1 so, 𝐿−1 { 2 } = 𝑒 2𝑡 𝑠𝑖𝑛 𝑠 2 +1 𝑠 −4𝑠+5 Question #12 find the Laplace transform of f (t). 𝑓(𝑡) = 𝑒 3𝑡 −1 1 Answer: Given The given function is 𝑓(𝑡) = formula used 𝑓(𝑡) 𝐿{ 𝑡 ∞ } = ∫𝑠 𝐹(𝜔)𝑑 𝜔 by using the above formula ∞ 𝑒 3𝑡 − 1 𝐿{ } = ∫ 𝐿{𝑒 3𝑡 − 1}𝑑𝜔 𝑡 𝑠 ∞ 1 1 =∫ { − }𝑑𝜔 𝜔−3 𝜔 𝑠 = [ln(𝜔 − 3) − ln(𝜔)]∞ 𝑠 𝑒 3𝑡 −1 1 = [ln ( 𝜔−3 𝜔 ∞ )] 𝑠 on further simplification, 𝑒 3𝑡 − 1 𝑠−3 𝐿{ } = 0 − ln ( ) 𝑡 𝑠 = −[ln(𝑠 − 3) − ln(𝑠)] = − ln(𝑠 − 3) + ln(𝑠) = ln(𝑠) − ln(𝑠 − 3) Question #13 Calculate the Laplace transform 𝐿𝑓 of the function 𝑓(𝑡) = 𝑒 4𝑡+2 − 8𝑡 6 + 8 sin(2𝑡) + 9 using the basic formulas and the linearity of the Laplace transform Answer: 𝑓(𝑡) = 𝑒 4𝑡+2 − 8𝑡 6 + 8 sin(2𝑡) + 9 𝐹(𝑡) = 𝑒 2 𝑒 4𝑡 − 8𝑡 6 + 8 sin(2𝑡) + 9 𝐿{𝑓(𝑡)} = 𝐿{𝑒 2 𝑒 4𝑡 − 8𝑡 6 + 8 sin(2𝑡) + 9} Apply linearity 𝐿{𝑓(𝑡)} = 𝑒 2 𝐿{𝑒 4𝑡 } − 8𝐿{𝑡 6 } − 8𝐿{sin(2𝑡)} + 9𝐿{1} using the basic formulas 1 1 𝐿{𝑒 𝑎𝑡 } = , 𝐿{1} = 𝑠−𝑎 𝑛! 𝐿{𝑡 𝑛 } = 𝑠 𝑠 𝑛+1 and 𝐿{sin(𝑎𝑡)} = So, 𝐿{𝑓(𝑡)} 𝑒2 𝐿{𝑓} = 𝑠−4 𝑎 𝑠 2 +𝑎2 6! 2 1 =𝑒 −8 7− 𝑠−4 𝑠 5760 16 9 − 𝑠7 − 𝑠 2 +22 + 8 2 𝑠 2 +22 +9 1 𝑠 𝑠 Question #14 𝑑 3 𝑥(𝑡) The Laplace transform of 𝐿{ 𝑑𝑡 3 } is given by Select one: sX(s)-x(0) 𝑠 3 𝑋(𝑠) − 𝑠 2 𝑥(0) − 𝑠𝑥̇ (0) − 𝑥̈ (0) 𝑠 3 𝑋(𝑠) − 𝑠 2 𝑥̈ (0) − 𝑠𝑥̇ (0) − 𝑥(0) 𝑠 2 𝑋(𝑠) − 𝑠𝑥(0) − 𝑥̇ (0) Answer: 𝑑 3 𝑥(𝑡) 𝐿{ } 𝑑𝑡 3 ′ (𝑡)] 𝐿[𝑥 ∞ −𝑠𝑡 )𝑥(𝑡)𝑑𝑡 = [𝑒 −𝑠𝑡 ⋅ 𝑥(𝑡)]∞ 0 − ∫0 (−𝑠𝑒 ∞ = −𝑥(0) + 𝑠 ∫0 𝑒 −𝑠𝑡 𝑥(𝑡)𝑑𝑡 (∵ 𝑒 −𝑠𝑡 𝑥(𝑡) = 0 when 𝑡 = ∞) = −𝑥(0) + 𝑠𝐿{𝑥(𝑡)} ∴ 𝐿{𝑥 ′ (𝑡)} = 𝑠𝐿[𝑥(𝑡)] − 𝑥(0) 𝐿{𝑥 ′′ (𝑡)} = 𝑠𝐿[𝑥 ′ (𝑡)] − 𝑥 ′ (0) = 𝑠{[𝑠𝑥(𝑠) − 𝑥(0)] − 𝑥 ′ (0)} = 𝑠 2 𝑥(𝑠) − 𝑠𝑥(0) − 𝑥 ′ (0) 𝐿{𝑥 ′′′ (𝑡)} = 𝑠 3 𝑥(𝑠) − 𝑠 2 𝑥(0) − 𝑠𝑥 ′ (0) − 𝑥 ′′ (0) 𝑑 3 𝑥(𝑡) 𝐿{ 𝑑𝑡 3 } = 𝑠 3 𝑥(𝑠) − 𝑠 2 𝑥(0) − 𝑠𝑥(0) − 𝑥(0) Question #15 determine the inverse Laplace transform of the function. 7 𝐿−1 {𝑅(𝑠)} = 𝐿−1 {(𝑠+3)(𝑠−3)} Answer: Step 1 7 Given function is, 𝑅(𝑠) = (𝑠+3)(𝑠−3) This can be re-written as, 7 𝑅(𝑠) = [ 1 6 𝑠−3 − 1 𝑠+3 ] Step 2 Inverse Laplace is given by, 1 1 𝐿−1 { } = 𝑒 −𝑎𝑡 and 𝐿−1 { } = 𝑒 𝑎𝑡 𝑠+𝑎 𝑠−𝑎 Step 3 Hence, Laplace inverse of the given function is, 7 1 𝐿−1 {𝑅(𝑠)} = [𝐿−1 {[ 7 = (𝑒 3𝑡 − 𝑒 6 6 −3𝑡 ) 𝑠−3 ]} − 𝐿−1 { 1 𝑠+3 }] Question #16 Find the solution of the given Initial Value Problem by using the Laplace Transform Method. 𝑦 ′′′ − 2𝑦 ′′ + 5𝑦 ′ = 𝑡, 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦 ′′ (0) = 2 Answer: Solution: 𝑦 ′′′ − 2𝑦 ′′ + 5𝑦 ′ = 𝑡, 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦 ′′ (0) = 2 𝐿{𝑦 ′′′ } = 𝑠 3 𝐿{𝑦} − 𝑠 2 𝑦(0) − 𝑠𝑦 ′ (0) − 𝑦 ′′ (0) 𝐿{𝑦 ′′ } = 𝑠 2 𝐿{𝑦} − 𝑠𝑦(0) − 𝑦 ′ (0) 𝐿{𝑦 ′ } = 𝑠𝐿{𝑦} − 𝑦(0) Apply Laplace transform we get [𝑠 3 𝐿{𝑦} − 5 − 2] − 2[𝑠 2 𝐿{𝑦} − 1] + 5[𝑠𝐿{𝑦}] = 𝐿{𝑡} 1 ∴ (𝑠 3 − 2𝑠 2 + 5𝑠)𝐿{𝑦} − 𝑠 = 2 ∴ (𝑠 3 2 − 2𝑠 + 5𝑠)𝐿{𝑦} = ∴ 𝐿{𝑦} = ∴ 𝐿{𝑦} = ∴ 𝐿{𝑦} = 1 + (𝑠2 𝑠 3 (𝑠 2 −2𝑠+5) −1 1 2 1 125 𝑠 −1 1 125 𝑠 + + + 25 𝑠 2 2 1 + 25 𝑠 2 𝑠 1 𝑠2 1 +𝑠 −2𝑠+5) 1 1 𝑠 + 5 𝑠3 1 1 + 5 𝑠3 1 125 (𝑠 2 −2𝑠+5) 𝑠 1 125 (𝑠 2 −2𝑠+5) − + 12 1 125 (𝑠 2 −2𝑠+5) 113 1 + (𝑠2 1 −2𝑠+5) 125 (𝑠 2 −2𝑠+5) ∴ Apply Inverse Laplace transform ∴𝑦= ∴𝑦= −1 125 −1 125 + + 2 25 2𝑡 25 𝑡+ + 𝑡2 10 𝑡2 10 + + 𝑒𝑡 sin(2𝑡) [ + cos(2𝑡)] + 125 2 𝑒 𝑡 cos(2𝑡) 114𝑒 𝑡 sin(2𝑡) 125 250 + 113 𝑒 𝑡 sin(2𝑡) 125 [ 2 ] This is required solution Question #17 Use the method of Laplace transformation to solve initial value problem. 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 = 𝑥 − 2𝑦, 𝑥(0) = −1, 𝑦(0) = 2 = 5𝑥 − 𝑦 Answer: We know in Laplace transformation 𝐿{𝑦 𝑛 (𝑡)} = 𝑠 𝑛 𝑦̅(𝑠) − 𝑠 𝑛−1 𝑦(0) − 𝑠 𝑛−2 𝑦 ′ (0) ⋯ 𝑦 𝑛−1 (0) 𝑛! 𝑠 𝑎 𝐿{𝑡 𝑛 } = 𝑛+1 , 𝐿{cos 𝑎 𝑡} = 2 , 𝐿{sin(𝑎𝑡)} = 2 2 𝑠 𝑠 +𝑎 𝑠 + 𝑎2 Now Given IVP is 𝑑𝑥 𝑑𝑡 𝑑𝑦 = 𝑥 − 2𝑦(1)𝑥(0) = −1, 𝑦(0) = 2 = 5𝑥 − 𝑦(2) 𝑑𝑡 ⇒ 𝑥 ′ (𝑡) − 𝑥 + 2𝑦 = 0 𝑦 ′ (𝑡) − 5𝑥 + 𝑦 = 0 Taking laplace transformation on both the equation and both the side From (1) 𝐿{𝑥(𝑡)} − 𝐿{𝑥(𝑡)} + 2𝐿{𝑦(𝑡)} = 0 ⇒ 𝑠𝑥(𝑠) − 𝑥(0) − 𝑥(𝑠) + 2𝑦(𝑠) = 0 ⇒ (𝑠 − 1)𝑥(𝑠) + 2𝑦(𝑠) = −1(3) From (2) 𝐿{𝑦 ′ (𝑡)} − 5𝐿{𝑥(𝑡)} + 𝐿{𝑦(𝑡)} = 0 ⇒ 𝑠𝑦(𝑠) − 𝑦(0) − 5𝑥(𝑠) + 𝑦(𝑠) = 0 ⇒ (𝑠 + 1)𝑦(𝑠) − 5𝑥(𝑠) = 2(4) ⇒ 5𝑥(𝑠 − 1)𝑥(𝑠) + 2𝑦(𝑠) = −1 (𝑠 − 1) × (𝑠 + 1)𝑦(𝑠) − 5𝑥(𝑠) = 2 ((𝑠 2 − 1) + 10)𝑦(𝑠) = 2(𝑠 − 1) − 5 2(𝑠−1)−5 ⇒ 𝑦(𝑠) = 𝑠 2 +9 =2 𝑠 𝑠 2 +9 − 7 3 3 𝑠 2 +9 Taking inverse laplace transformation both the side. We get 𝑠 7 3 𝑦(𝑡) = 2𝐿−1 { 2 } − 𝐿{ 2 } 𝑠 +9 3 𝑠 +9 7 ⇒ 𝑦(𝑡) = 2 cos 3 𝑡 − sin 3 𝑡 3 From equation (2) 5𝑥(𝑡) = 𝑦 ′ (𝑡) + 𝑦(𝑡) 1 7 5 1 −25 3 ⇒ 𝑥(𝑡) = (−6 sin 3 𝑡 − 7 cos 3 𝑡 + 2 cos 3 𝑡 − sin 3 𝑡) ⇒ 𝑥(𝑡) = ( sin 3 𝑡 − 5 cos 3 𝑡) 5 3 5 ⇒ 𝑥(𝑡) = − sin 3 𝑡 − cos 3 𝑡 3 Question #18 Find the Laplace transform of the following: ℎ(𝑡) = 𝑡 2 𝑒 −𝑡 ℎ(𝑡 − 2) Answer: 𝐿{𝑒 −𝑡 𝑡 2 𝐻(𝑡 − 2)} Use laplace transform table: 𝐿{𝑡 𝑘 𝑓(𝑡)} = (−1)𝑘 2 −𝑡 −𝑡 For 𝑡 𝑒 𝐻(𝑡 − 2): 𝑓(𝑡) = 𝑒 𝐻(𝑡 − 2), 𝑘 = 2 = (−1)2 𝑑2 𝑑𝑠 2 (𝐿{𝑒 −𝑡 𝐻(𝑡 − 2)}) 𝐿{𝑒 −𝑡 𝐻(𝑡 − 2)}: 𝑑2 𝑑𝑠 2 = ( 𝑒 −2(𝑠+1) 𝑠+1 )= 𝑒 −2(𝑠+1) 𝑠+1 2𝑒 −2(𝑠+1) (2𝑠 2 +6𝑠+5) (𝑠+1)3 2𝑒 −2(𝑠+1) (2𝑠 2 +6𝑠+5) (−1)2 (𝑠+1)3 Question #19 𝑑𝑘 𝑑𝑠 𝑘 (𝐿{𝑓(𝑡)}) Find the inverse Laplace transform of the following: 2𝑒 −𝑠 𝐻(𝑠) = 𝑠 2 (𝑠+3) Answer: 𝐿−1 { 2𝑒 −𝑠 } 𝑠 2 (𝑠+3) Apply inverse transform rule : if 𝐿−1 {𝐹(𝑠)} = 𝑓(𝑡) then 𝐿−1 {𝑒 −𝑎𝑠 𝐹(𝑠)} = 𝐻(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) Where H(t) is Heaviside step function 2𝑒 −𝑠 For 𝑠 2 (𝑠+3) = 𝐻(𝑡 − 1)𝐿 { 2 𝑠 2 (𝑠+3) 2 −1 𝐿−1 { 2 : 𝐹(𝑠) = }(𝑡 − 1) 𝑠 2 (𝑠+3) 2 2𝑡 9 2 3 }: − 𝐻(𝑡) + 𝑠 2 (𝑠+3) ,𝑎 = 1 2 + 𝑒 −3𝑡 = 𝐻(𝑡 − 1) (− 𝐻(𝑡 − 1) + 9 2(𝑡−1) 9 3 2 + 𝑒 −3(𝑡−1) ) 9 Question #20 Use the table of laplace transform to find the inverse Laplace transform of 𝐹(𝑠) = Answer: Step 1 solution: We need to find the inverse laplace transform of 𝐹(𝑠) = 2+𝑠(𝑠+1) 𝑠(𝑠 2 −𝑠−6) Step 2 For this 𝐹(𝑠) = = 2+𝑠(𝑠+1) 𝑠(𝑠 2 −𝑠−6) 2 𝑠+1 𝑠(𝑠 2 −𝑠−6) 𝐹(𝑠) = + (𝑠2 −𝑠−6) 2 𝑠+1 + (𝑠−3)(𝑠+2) 𝑠(𝑠−3)(𝑠+2) using partial fraction 1 2 1 1 1 4 1 𝐹(𝑠) = − + + (𝑠+2) + (𝑠−3) + 3𝑠 15 𝑠−3 5 1 5 5(𝑠+2) Taking inverse laplace transform is 1 1 2 𝑓(𝑡) = 𝐿−1 {𝐹(𝑠)} = − 𝐿−1 { } + 𝐿−1 { 1 𝑓(𝑡) = − + 2𝑒 3𝑡 3 𝑓(𝑡) = 3 + 𝑒 −2𝑡 𝑠 4 15 𝑒 −2𝑡 + 𝑒 3𝑡 + 15 5 5 14 3𝑡 2 1 𝑒 + 𝑒 −2𝑡 − 15 5 3 1 𝑠−3 5 Hence, inverse laplace transform of F(s) 14 2 1 𝑓(𝑡) = 𝐿−1 {𝐹(𝑠)} = 𝑒 3𝑡 + 𝑒 −2𝑡 − ⇒ 𝐿−1 {𝐹(𝑠)} = 14 15 15 2 5 𝑒 3𝑡 + 𝑒 −2𝑡 − 5 1 3 3 1 1 5 𝑠+2 } + 𝐿−1 { 4 1 5 𝑠−3 } + 𝐿−1 { 1 1 5 𝑠+2 } + 𝐿−1 { } 2+𝑠(𝑠+1) 𝑠(𝑠 2 −𝑠−6)