Questions and Answers #5 Laplace Transform

Questions and Answers Sheet 5 Laplace Transform Question #1 Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform 𝐿{𝑡 4 + 𝑡 3 − 𝑡 2 − 𝑡 + sin √2 𝑡} Answer: The given Laplace transform is 𝐿{𝑡 4 + 𝑡 3 − 𝑡 2 − 𝑡 + sin √2 𝑡} By using the linearity property of Laplace transform For Function f(t), g(t) and constant a,b: 𝐿{𝑎𝑓(𝑡) + 𝑏𝑔(𝑡)} = 𝑎𝐿{𝑓(𝑡)} + 𝑏𝐿{𝑔(𝐻)} So from eqn (1) 𝐿{𝑡 4 + 𝑡 3 − 𝑡 2 − 𝑡 + sin √2 𝑡} = 𝐿{𝑡 4 } + {𝑡 3 } − {𝑡 2 } − {𝑡} + {sin √2 𝑡}(2) By using 𝐿{𝑡 𝑛 } = 𝑎 𝐿{sin(𝑎𝑡)} = 𝑛! 𝑠 𝑛+1 and we will get from eqn (2) 𝑠 2 +𝑎2 𝐿{𝑡 4 + 𝑡 3 − 𝑡 2 − 𝑡 + sin √2 𝑡} = 𝐿{𝑡 4 + 𝑡 3 − 𝑡 2 − 𝑡 + sin √2 𝑡} = 4! 𝑠5 4! 𝑠5 + + 3! 𝑠4 3! 𝑠4 − − 2! 𝑠3 2! 𝑠3 − − 1 𝑠2 1 𝑠2 + + √2 2 𝑠 2 +(√2) √2 2 𝑠 2 +(√2) Question #2 Find the inverse Laplace transform of the following 𝑠+4 𝑠 2 +4𝑠+8 Answer: 𝑠+4 𝑠+4 𝑠+4 = = 𝑠 2 + 4𝑠 + 8 𝑠 2 + 4𝑠 + 4 + 4 (𝑠 + 2)2 + 22 𝑠+2 2 = (𝑠+2)2 2 + (𝑠+2)2 2 +2 −1 Now, 𝐿 { +2 𝑠+4 𝑠+2 } = 𝐿−1 {(𝑠+2)2 𝑠 2 +4𝑠+8 −2𝑡 −2𝑡 +22 2 } + 𝐿−1 {(𝑠+2)2 +22 } = cos(2𝑡) 𝑒 +𝑒 sin(2𝑡) = (cos(2𝑡) + sin(2𝑡))𝑒 −2𝑡 Question #3 Find the inverse Laplace transform of the following: 7 𝑠−8 2 𝑠+8 ( ) log ( ) Answer: Given: 7 𝑠−8 2 𝑠+8 ( ) log ( ) To find: The inverse Laplace transform. 7 𝑠−8 2 𝑠+8 Let 𝑓(𝑠) = ( ) log 𝑒 ( −1 𝐹(𝑡) = 𝐿 {𝑓(𝑠)} 7 𝑠−8 2 𝑠+8 𝑓(𝑠) = ( ) log 𝑒 ) 7 = [log 𝑒 (𝑠 − 8) − log 𝑒 (𝑠 + 8)] 2 7 𝑓 ′ (𝑠) = [ 1 1 − 2 𝑠−8 𝑠+8 ] Differentaking f(s) with respect to s is eqvivalent to multiplying F(t) by t ∴ 𝐿{𝐹(𝑡)} = − 𝑑𝑓(𝑠) 𝑑𝑠 =-f'(s) ∴ 𝑡𝐹(𝑡) = −𝐿−1 {𝑓 ′ (𝑠)} 7 1 2 7 𝑠−8 = − 𝐿−1 { =− 2 [𝑒 8𝑡 = −7 ( − −𝑒 1 𝑠+8 −8𝑡 ] 𝑒 8𝑡 −𝑒 −8𝑡 2 } ) = −7 sin ℎ 8𝑡 (∵ ℎ𝑥 = 𝑒 𝑥 −𝑒 −𝑥 ∴ 𝐹(𝑡) = ) 2 −7 sin 8𝑡 𝑡 Question #4 Find the Laplace Transform of the function. 𝑥(𝑡) = 3𝑒 4(𝑡+2) 𝑢(𝑡 + 2) + 3(𝑡 − 5)𝑢(𝑡 − 5) + 3 sin(𝑡 + 2) 𝑢(𝑡 + 2) Answer: Step 1 Given: 𝑥(𝑡) = 3𝑒 4(𝑡+2) 𝑢(𝑡 + 2) + 3(𝑡 − 5)𝑢(𝑡 − 5) + 3 sin(𝑡 + 2) 𝑢(𝑡 + 2) To find the Laplace transform of the function. Step 2 Now 𝐿{𝑥(𝑡)} = 3𝐿{𝑒 4(𝑡+2) 𝑢(𝑡 + 2)} + 3𝐿{(𝑡 − 5)𝑢(𝑡 − 5)} + 3𝐿{sin(𝑡 + 2) 𝑢(𝑡 + 2)} Take 𝐿{𝑒 4(𝑡+2) 𝑢(𝑡 + 2)} = 𝐿{𝑒 4(𝑡(−2)) 𝑢(𝑡 − (−2))} we have 𝐿{𝑓(𝑡 − 𝑎)𝑢(𝑡 − 𝑎)} = 𝑒 −𝑎𝑠 𝐹(𝑠) Here 𝑓(𝑡 − (−2)) = 𝑒 4(𝑡−(−2)) ⇒ 𝑓(𝑡 + 2) = 𝑒 4(𝑡+2) put t=t-2 ⇒ 𝑓(𝑡) = 𝑒 4𝑡 ⇒ 𝐿{𝑓(𝑡)} = 𝐿{𝑒 4𝑡 } 1 = = 𝐹(𝑠) 𝑠−4 1 ∴ 𝐿{𝑒 4(𝑡+2) 𝑢(𝑡 + 2)} = 𝑒 2𝑠 ⋅ 𝑠−4 Take 𝐿{(𝑡 − 5)𝑢(𝑡 − 5)} here, f(t-5)=t-5 ⇒ 𝑓(𝑡) = 𝑡 1 ⇒ 𝐿{𝑓(𝑡)} = 𝐿{𝑡} = 2 = 𝐹(𝑠) 𝑠 1 ∴ 𝐿{(𝑡 − 5)𝑢(𝑡 − 5)} = 𝑒 −5𝑠 2 𝑠 Take 𝐿{sin(𝑡 + 2) 𝑢(𝑡 + 2)} Here , 𝑓(𝑡 + 2) = sin(𝑡 + 2) ⇒ 𝑓(𝑡) = 𝑠𝑖𝑛 ⇒ 𝐿{𝑓(𝑡)} = 𝐿{sin 𝑡} 𝑎 we have 𝐿 sin 𝑎 𝑡 = 2 2 = 1 𝑠 2 +1 𝑠 +𝑎 = 𝐹(𝑠) ∴ 𝐿{sin(𝑡 + 2) ⋅ 𝑢(𝑡 + 2)} = 𝑒 2𝑠 1 𝑠 2 +1 From (1) = 7 × (𝑠) = 3𝑒 2𝑠 = 7 × (𝑠) = 3𝑒 2𝑠 [ 1 𝑠−4 + 1 𝑠 2 +1 1 𝑠−4 ]+ + 3𝑒 −5𝑠 3𝑒 −5𝑠 1 𝑠2 + 3𝑒 2𝑠 1 𝑠 2 +1 𝑠2 Question #5 Find the Laplace transform of cos 2 Answer: Step 1 Solution The laplace transform of 𝑠 (1) cos(𝑎𝑡) ⇒ 2 𝑠 + 𝑎2 and for some constant k 𝑘 𝑘 ⇒ Laplace (2) 𝑠 Step 2 Let 𝑓(𝑡) = cos 2 (𝑡), Let say F(s) is laplace transform of f(t) 𝑓(𝑡) = cos 2 (𝑡) 1 cos(2𝑡) 2 2 2 ⇒ 𝑓(𝑡) = + (3) [cos(2𝜎) = 2 cos 𝜎 − 1] Take Laplace trasform of f(t) as per equation (3) From 1 and 2 1 1 𝑠 𝐹(𝑠) = + × 2 2 2𝑠 2 𝑠 +2 1 𝑠 (4) 𝐹(𝑠) = + 2 2𝑠 2(𝑠 + 4) Question #6 Find the Laplace transform of 3𝑒 −7𝑡 Answer: As: 𝐹(𝑥) = 3𝑒 −7𝑡 ∵ 𝐿{𝑒 𝑎𝑡 } = 1 𝑠−𝑎 𝐿{𝑓(𝑥)} = 3𝐿{𝑒 −7𝑡 } put a=-7 1 𝐿{𝑓(𝑥)} = 3 × = 𝑠−(−7) 3 𝑠+7 −7𝑡 Result: 𝐿{𝐹(𝑥)} = 𝐿{3𝑒 }= 3 𝑠+7 Question #7 Determine the Laplace transform of the following integrals 𝑡 𝐿{∫0 𝑒 −2𝑡 𝑡 2 𝑑𝑡} Answer: 𝑡 𝐿{∫0 𝑒 −2𝑡 𝑡 2 𝑑𝑡} We know that , if 𝐿[𝑓(𝑡)] = 𝐹(𝑠) 𝑡 then 𝐿 [∫0 𝑓(𝑣)𝑑𝑣 ] = Here we have 𝐹(𝑠) 𝑠 𝑡 ∫0 𝑒 −2𝑡 𝑡 2 𝑑𝑡 so 𝑓(𝑡) = 𝑒 −2𝑡 𝑡 2 Since, if 𝐿[𝑔(𝑡)] = 𝐺(𝑠) then 𝐿[𝑒 𝑎𝑡 𝑔(𝑡)] = 𝐺(𝑠 − 𝑎) so it we take 𝑔(𝑡) = 𝑡 2 2 ⇒ 𝐺(𝑠) = 𝐿[𝑡 2 ] = 3 𝑠 and hence 2 𝐿[𝑒 −2𝑡 𝑡 2 ] = ( 3 ) 𝑠 𝑠⇒𝑠−(−2) 2 ⇒ 𝐿[𝑒 −2𝑡 𝑡 2 ] = (𝑠+2)3 Then 𝑡 𝐿 [∫0 𝑒 −2𝑡 𝑡 2 𝑑𝑡] = ⇒ 𝑡 𝐿 [∫0 𝑒 −2𝑡 𝑡 2 𝑑𝑡] 2 (𝑠+2)3 𝑠 2 = 𝑠(𝑠+2)3 Question #8 Without using the shift formula prove that Laplace of 𝑓(𝑡) = 𝑒 −𝑎𝑡 cos 𝑤 𝑡𝑎𝑠𝑘 𝑖𝑠 𝑤2 (𝑠+𝛼)2 +𝑤 2 Answer: Step 1 Given, 𝑓(𝑡) = 𝑒 −𝑎𝑡 cos 𝑤 𝑡 𝑠+𝛼 To prove that, 𝐿(𝑓(𝑡)) = (𝑠+𝛼)2 By definition, 𝐿(𝑓(𝑡)) = = ∞ ∫0 𝑒 −(𝛼+𝑠)𝑡 +𝑤2 ∞ −𝑠𝑡 −𝑎𝑡 ∫0 𝑒 𝑒 cos 𝑤 𝑡𝑑𝑡 cos 𝑤 𝑡𝑑𝑡 Step 2 ∞ Let 𝐼 = ∫0 𝑒 −(𝛼+𝑠)𝑡 cos 𝑤 𝑡𝑑𝑡 = [cos 𝑤 𝑡 × = = = 1 𝑒 −(𝛼+𝑠)𝑡 −(𝛼+𝑠) ∞ 0 𝑤 𝛼+𝑠 ∞ ] − ∫0 −𝑤 sin 𝑤 𝑡 × − (𝛼+𝑠) {[sin 𝑤 𝑡 × 𝑒 −(𝛼+𝑠)𝑡 ∞ 𝑒 −(𝛼+𝑠)𝑡 −(𝛼+𝑠) ∞ ] − ∫0 𝑤 cos 𝑤 𝑡 × −(𝛼+𝑠) 0 ∞ −(𝛼+𝑠)𝑡 1 𝑤 𝑤 − [ ∫ 𝑒 𝛼 + 𝑠 (𝛼 + 𝑠) (𝛼 + 𝑠) 0 cos 𝑤 𝑡𝑑𝑡] 𝑤2 1 𝛼+𝑠 − (𝛼+𝑠)2 𝐼 Therefore, the Laplace transform is, 𝐿(𝑓(𝑡)) = 1 𝑠+𝛼 (𝑠+𝛼)2 [(𝑠+𝛼)2 +𝑤2 ] 𝑠+𝛼 = (𝑠+𝛼)2 +𝑤2 Question #9 Solve using the inverse Laplace transform 5 𝐹(𝑠) = 2 𝑠 +𝑠−6 Answer: Answer: Given: 𝐹(𝑠) = 5 𝑠 2 +𝑠−6 𝑑𝑡 𝑒 −(𝛼+𝑠)𝑡 −(𝛼+𝑠) 𝑑𝑡} 5 𝐹(𝑠) = 1 2 𝑠 2 +2× 𝑠+0.52 −0.52 −6 5 = (𝑠+0.5)2 −6⋅25 5 𝐹(𝑠) = (𝑠+0.5)2 −2⋅52 Apply inverse Laplace transform to F(s) 5 𝐿−1 𝐿−1 (𝑠+0.5)2 2 −2⋅5 𝑏 −1 Note: 𝐿 (𝑠+𝑎)2 −𝑏 2 −0.5𝑡 = 𝑒 −𝑎𝑡 ⋅ sin ℎ (𝑏𝑡) ∴ 𝑓(𝑡) = 2 × 𝑒 sin ℎ (2 ⋅ 5𝑡) 𝑓(𝑡) = 2 × 𝑒 −0.5𝑡 sin ℎ (2 ⋅ 5𝑡) Question #10 Laplace transform which one is correct 𝐿{𝑡 + 𝑒 3𝑡 } 1 1 1) 𝐿{𝑡 + 𝑒 3𝑡 } = 2 + 2) 𝐿{𝑡 + 𝑒 3𝑡 } = 3) 𝐿{𝑡 + 𝑒 3𝑡 } = 4) 𝐿{𝑡 + 𝑒 3𝑡 } = 𝑠 1 𝑠−3 1 𝑠 1 𝑠−3 1 2 − 2 + 𝑠 1 − 𝑠2 𝑠+3 1 𝑠+3 Answer: Step 1 Find the laplace transform of 𝐿{𝑡 + 𝑒 3𝑡 } definition of Laplace Transform of f(t) ∞ 𝐿{𝑓(𝑡)} = ∫0 𝑒 −𝑠𝑡 𝑓(𝑡)𝑑𝑡 Step 2 ∞ So 𝐿{𝑡 + 𝑒 3𝑡 } = ∫0 𝑒 −𝑠𝑡 (𝑡 + 𝑒 3𝑡 )𝑑𝑡 ∞ ∞ = ∫0 𝑡𝑒 −𝑠𝑡 𝑑𝑡 + ∫0 𝑒 −(𝑠−3)𝑡 𝑑𝑡 = [𝑡 𝑒 −𝑠𝑡 −𝑠 − 𝑒 −𝑠𝑡 𝑠2 ∞ 𝑒 −(𝑠−3)𝑡 ] +[ 0 −(𝑠−3) 1 0 1 𝑠 𝑠−3 = (0 + 0 + 0 + 2) + (0 + ∵𝑒 = −∞ 1 𝑠2 ∞ ] ) =0 + 1 𝑠−3 3𝑡 𝐿{𝑡 + 𝑒 } = 1 𝑠2 + 1 𝑠−3 Question #11 Find the Laplace transform of equation. 5 𝑑 3 𝑥(𝑡) 𝑑𝑡 3 𝑑 2 𝑥(0) 𝑑𝑡 2 − 8 𝑑 2 𝑥(𝑡) 14 𝑑𝑡 2 𝑑𝑥(0) = 1, 𝑑𝑡 −9 𝑑𝑥(𝑡) 𝑑𝑡 + 𝑥(𝑡) = 𝑢(𝑡) = −2, 𝑥(0) = 1 Answer: Step 1 5 𝑑 3 𝑥(𝑡) 𝑑𝑡 3 − 8 𝑑 2 𝑥(𝑡) 14 𝑑𝑡 2 −9 𝑑𝑥(𝑡) 𝑑𝑡 + 𝑥(𝑡) = 𝑢(𝑡) ⇒ Taking Laplace transform of both sides 𝐿{5 𝑑 3 (𝑥) 𝑑𝑡 3 − 8 𝑑 2 (𝑥) 14 𝑑𝑡 2 −9 𝑑𝑥 𝑑𝑡 + 𝑥} = 𝐿{𝑢(𝑡)} 8 1 ⇒ {5[𝑠 3 𝑥̅ − 𝑠 2 𝑥(0) − 𝑠(𝑥 ′ (0)) − 𝑥 ′′ ] − ( ) [𝑠 2 𝑥̅ − 𝑠𝑥(0) − 𝑥 ′ (0)] − 9[𝑠𝑥̅ − 𝑥(0)] + 𝑥̅ } = ( ) 14 8 3 ⇒ 𝑥̅ (5𝑠 − 14 2 2 𝑠 − 9𝑠 + 1) = [5𝑠 − 10𝑠 − 𝑠 8 14 8 1 7 𝑠 +5+ −9+ ] Step 2 Simplify further, we get ∵ Laplace transform of given equation is 8 148 14 14 𝑥̅ (5𝑠 3 − ( ) 𝑠 2 − 9𝑠 + 1) = (5𝑠 2 − ( 20 1 7 𝑠 )𝑠 − ( ) + ) Question #12 Find the laplace transform of this function 𝑓(𝑡) = 𝑒 −2𝑡 sin 3 𝑡 𝑠𝑖𝑛 Answer: 𝑓(𝑡) = 𝑒 −2𝑡 sin 3 𝑡 sin 𝑓(𝑡) = 𝑒 −2𝑡 sin 3 𝑡 sin 𝑓(𝑡) = 𝑒 −2𝑡 sin 3 𝑡 sin 𝑒 −2𝑡 [cos(3𝑡 − 𝑡) − cos(3𝑡 + 𝑡)] = 2 −2𝑡 𝑒 [cos 2 𝑡 − cos 4 𝑡] = 2 [sin 𝛼 sin 𝛽 = 1 [cos(𝛼 − 𝛽) − cos(𝛼 + 𝛽)]] 2 𝑒 −2𝑡 𝐿{𝑓(𝑡): 𝑠} = 𝐿{ 2 1 [cos 2 𝑡 − cos 4 𝑡]: 𝑠} = 𝐿{(cos 2 𝑡 − cos 4 𝑡): 𝑠 + 2} 2 [𝐿{𝑒 −𝑎𝑡 𝑓(𝑡): 𝑠} = 𝐿{𝑓(𝑡): 𝑠 + 𝑎}] 1 = [𝐿{cos 2 𝑡: 𝑠 + 2} − 𝐿{cos 4 𝑡: 𝑠 + 2}] 2 ∞ 𝐿[𝑒 𝑎𝑥 : 𝑠] = ∫0 𝑒 𝑎𝑥 𝑒 −𝑠𝑥 𝑑𝑥 ∞ = ∫0 𝑒 −(𝑠−𝑎)𝑥 𝑑𝑥 =[ 𝑒 −(𝑠−𝑎)𝑥 −(𝑠−𝑎) 1 = ∞ ] 0 for 𝑠 > 𝑎 𝑠−𝑎 [∵ lim 𝑒 −(𝑠−𝑎)𝑥 = 0] 𝑥⇒∞ so 𝐿{𝑒 𝑖𝜔𝑥 : 𝑠} = 1 𝑠−𝑖𝜔 𝑠+𝑖𝜔 = (𝑠−𝑖𝜔)(𝑠+𝑖𝜔) 𝑠+𝑖𝜔 = 𝑠 2 +𝜔2 ⇒ 𝐿{cos(𝜔𝑤) + 𝑖 sin(𝜔𝑥) : 𝑠} = 𝑠 𝑠 2 +𝜔2 𝜔 +𝑖 𝑠 2 +𝜔2 change variable x=t and compary real part, we get 𝑠 𝐿{cos 𝜔 𝑡: 𝑠} = 2 2 𝑠 +𝜔 𝑒 −2𝑡 𝐿{ 2 [cos 2 𝑡 − cos 4 𝑡]: 𝑠} = 1 𝑠+2 2 (𝑠+2)2 +4 − 1 𝑠+2 2 (𝑠+2)2 +16 Question #13 Solve equation using Laplace transform 𝜋 𝑦̈ + 9𝑦 = 𝛿 (𝑡 − ) sin(𝑡) , 𝑦(0) = 0, 𝑦̇ (0) = 0 6 Answer: Step 1 The given equation is 𝜋 𝑦̈ + 9𝑦 = 𝛿 (𝑡 − ) sin(𝑡) , 𝑦(0) = 0, 𝑦̇ (0) = 0 6 Use the property of delta function. 𝜋 𝑦̈ + 9𝑦 = 𝛿 (𝑡 − ) sin(𝑡) , 𝑦(0) = 0, 𝑦̇ (0) = 0 6 ∞ ∫−∞ 𝛿(𝑡 − 𝑡 ′ )𝜙(𝑡)𝑑𝑡 = 𝜙(𝑡 ′ ) Step 2 Calculation 𝜋 𝑦 ′′ + 9𝑦 = 𝛿 (𝑡 − ) sin(𝑡) , 𝑦(0) = 0, 𝑦̇ (0) = 0 6 Take the Laplace in both sides. 𝜋 𝐿[𝑦 ′′ ] + 9𝐿[𝑦] = 𝐿 [𝛿 (𝑡 − ) sin(𝑡)] 6 𝜋 [𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0)] + 9𝑌(𝑠) = 𝐿 [𝛿 (𝑡 − ) sin(𝑡)] 6 ∞ ∫0 [𝛿 (𝑡 𝜋 𝐿 [𝛿 (𝑡 − ) sin(𝑡)] = 6 𝜋 − )𝑒 6 1 −𝑠𝑡 𝜋 𝜋 1 𝜋𝑠 sin(𝑡)] 𝑑𝑡 = sin ( ) 𝑒 −𝑠 6 = 𝑒 − 6 6 𝜋𝑠 2 ⇒ [𝑠 2 𝑌(𝑠) − 0 − 0] + 9𝑌(𝑠) = 𝑒 − 6 2 𝜋𝑠 1 ⇒ 𝑠 2 𝑌(𝑠) + 9𝑌(𝑠) = 𝑒 − 6 1 ⇒ 𝑌(𝑠)(𝑠 2 + 9) = 𝑒 2 𝜋𝑠 − 6 2 ⇒ 𝜋𝑠 1 − 𝑒 6 𝑌(𝑠)(𝑠 2 +9) 2 (𝑠 2 +9) (𝑠 2 +9) 𝜋𝑠 −6 𝑒 = ⇒ 𝑌(𝑠) = 2(𝑠 2 +9) 1 Let 𝐻(𝑠) = 2(𝑠 2 +9) −1 ℎ(𝑡) = 𝐿 {𝐻} = 𝐿−1 { 1 }= sin(3𝑡) 2(𝑠 2 +9) 2⋅3 = sin(3𝑡) 6 Apply t-shifting theorem 𝐿−1 [𝑒 −𝑎𝑠 𝐹(𝑠)] = 𝑓(𝑡 − 𝑎)𝑢(𝑡 − 𝑎) 𝜋𝑠 1 𝜋 𝜋 6 𝜋 6 6 𝐿−1 [𝑌(𝑠)] = 𝐿−1 [𝑒 − 6 𝐻(𝑠)] = sin [3 (𝑡 − )] 𝑢 (𝑡 − ) 1 𝜋 ⇒ 𝑦(𝑡) = sin (3𝑡 − ) 𝑢 (𝑡 − ) 6 2 6 1 𝜋 6 6 ⇒ 𝑦(𝑡) = − cos(3𝑡) 𝑢 (𝑡 − ) 1 𝜋 6 6 Hence the Laplace transform of the given equation is − cos(3𝑡) 𝑢 (𝑡 − ) Question #14 Determine the Laplace transform of the following functions. 2 sin 3 𝑎𝑡 where a is any positive constant and s>a 𝑡 Answer: Step 1 Given: The function is 2 𝑡 sin(3𝑎𝑡), where a is any positive constant and s>a Step 2 Let the function 2 𝑓(𝑡) = sin(3𝑎𝑡) 𝑡 1 ∞ if 𝐿[𝑓(𝑡)] = 𝐹(𝑠),then 𝐿 [ 𝑓(𝑡)] = ∫𝑠 𝐹(𝑠)𝑑𝑠 𝑡 We know that 𝐿[sin(𝑎𝑡)] = 3𝑎 Then 𝐿[sin(3𝑎𝑡)] = 𝑎 𝑠 2 +𝑎2 𝑠 2 +(3𝑎)2 3𝑎 ⇒ 𝐿[sin(3𝑎𝑡)] = 𝑠 2 +9𝑎2 Step 3 2 1 Then 𝐿 [ sin(3𝑎𝑡)] = 2𝐿 [ sin(3𝑎𝑡)] 𝑡 𝑡 ∞ 1 Since 𝐿 [ 𝑓(𝑡)] = ∫𝑠 𝐹(𝑠)𝑑𝑠 𝑡 ∞ 1 3𝑎 2𝐿 [ sin(3𝑎𝑡)] = 2 ∫ 2 𝑑𝑠 𝑡 𝑠 + 9𝑎2 0 ∞ 2 𝐿 [ sin(3𝑎𝑡)] = 6𝑎 ∫0 𝑡 1 𝑠 2 +(3𝑎)2 𝑑𝑠 2 1 1 ∞ 𝐿 [ sin(3𝑎𝑡)] = 6𝑎 [ tan−1 ( )] 𝑡 3𝑎 3𝑎 0 2 𝜋 𝐿 [ sin(3𝑎𝑡)] = 2 [ − 0] 𝑡 2 2 Therefore, 𝐿 [ sin(3𝑎𝑡)] = 𝜋 𝑡 Step 4 Answer: 2 𝐿 [ sin(3𝑎𝑡)] = 𝜋 𝑡 Question #15 Find the INVERSE Laplace transform: 𝐹(𝑠) = 2𝑠+16 𝑠 2 +4𝑠+19 Answer: Step 1 Given, 𝐹(𝑠) = 2𝑠+16 𝑠 2 +4𝑠+19 This transform can be written as, 2𝑠+16 𝐹(𝑠) = 2 𝑠 +4𝑠+19 2𝑠+16 = (𝑠+2)2 +15 2𝑠+4 = (𝑠+2)2 +15 2(𝑠+2) = (𝑠+2)2 +15 12 + (𝑠+2)2 +15 12 + (𝑠+2)2 +15 Step 2 Taking inverse Laplace transform of both sides, 2(𝑠+2) 𝑓(𝑡) = 𝐿−1 [(𝑠+2)2 12 +15 (𝑠+2) = 2𝐿−1 [ 2 (𝑠+2)2 +(√15) ] + 𝐿−1 [(𝑠+2)2 ] + 12𝐿−1 [ +15 ] 1 2 (𝑠+2)2 +(√15) ] Using the formula, 𝐿−1 [𝐹(𝑠)] = 𝑓(𝑡) ⇒ 𝐿−1 [𝐹(𝑠 − 𝑎)] = 𝑒 𝑎𝑡 𝑓(𝑡) 𝐿−1 [ 𝑠 ] = cos(𝑎𝑡) 𝑠 2 +𝑎2 𝑎 −1 𝐿 [ 2 2] 𝑠 +𝑎 Step 3 = sin(𝑎𝑡) Therefore, 𝑓(𝑡) = 2𝐿−1 [ (𝑠+2) 2 (𝑠+2)2 +(√15) −2𝑡 1 = 2 × 𝑒 −2𝑡 cos(√15𝑡) + 12 × 𝑒 ]+ √15 12 √15 𝐿−1 [ √15 2 (𝑠+2)2 +(√15) ] sin(√15𝑡) Hence, inverse Laplace transform of the given function is 𝑓(𝑡) = 2𝑒 −2𝑡 cos(√15𝑡) + Question #16 determine the inverse Laplace transform of the given function. 2𝑠 + 1 𝐹(𝑠) = 2 𝑠 + 16 Answer: Step 1 Find the inverse Laplace transform of 𝐹(𝑠) = 2𝑠+1 𝑠 2 +16 Step 2 𝐿−1 { = 2𝑠+1 2𝑠 1 } + 𝐿−1 { 2 } 𝑠 2 +16 𝑠 +16 1 + 𝐿−1 { 2 } 𝑠 +16 1 𝑠 + sin 4 𝑡 [∵ 𝐿−1 { 2 2 } = 4 𝑠 +𝑎 } = 𝐿−1 { 𝑠 2 +16 𝑠 2𝐿−1 { 2 } 𝑠 +16 = 2(cos 4 𝑡) cos 𝑎 𝑡 and 𝐿−1 { 𝑎 𝑠 2 +𝑎2 } = sin 𝑎 𝑡] 1 = 2 cos 4 𝑡 + sin 4 𝑡 4 Question #17 determine the inverse Laplace transform of the given function. 2𝑠 𝐹(𝑠) = 2 𝑠 +9 Answer: Step 1 Given 𝐹(𝑠) = 2𝑠 𝑠 2 +9 Step 2 𝐿−1 { 𝑠 𝑠 2 +𝑎2 so 𝐿−1 { = = 2𝑠 } = cos 𝑎 𝑡 } 𝑠 2 +9 𝑠 2𝐿−1 { 2 } 𝑠 +9 𝑠 2𝐿−1 { 2 } 𝑠 +9 𝑠 −1 = 2𝐿 { 𝑠 2 +32 } = 2 cos 3 𝑡 Question #18 determine the inverse Laplace transform of the given function. 𝑠−2 𝐹(𝑠) = (𝑠+1)(𝑠2 Answer: Step 1 +4) 12 √15 𝑒 −2𝑡 sin(√15𝑡) 𝑠−2 Given Laplace transform is 𝐹(𝑠) = (𝑠+1)(𝑠2 +4) Step 2 Obtain the inverse Laplace transform 𝑠−2 𝐿−1 {𝐹(𝑠)} = 𝐿−1 {(𝑠+1)(𝑠2 } +4) 3 = 𝐿−1 {− 5(𝑠+1) 3 −1 1 =− 𝐿 { = = 3𝑠 + } 5(𝑠 2 +4) 5(𝑠 2 +4) 3 −1 𝑠 2 2 𝐿 {(𝑠2 } + 𝐿−1 {(𝑠2 } 5 +4) 5 +4) 2 1 }+ 5 3 2 + 𝑠+1 3 −𝑡 − 𝑒 + cos(2𝑡) + ( sin(2𝑡)) 5 5 5 2 3 −𝑡 3 cos(2𝑡) sin(2𝑡) − 𝑒 + + 5 5 5 Question #19 Evaluate the laplace transform for this a) 𝐿{4𝑥 2 − 3 cos(2𝑥) + 5𝑒 𝑥 } 1 b) 𝐿{4 + sin(𝑥) − 𝑒 4𝑥 } 2 Answer: Step 1 Introduction Use following laplace formulas in solving following problems 1 1) 𝐿{1} = 𝑠 2) 𝐿{𝑥 𝑛 } = 𝑛! 𝑠 𝑛+1 𝑠 3) 𝐿{cos 𝑎 𝑥} = 4) 𝐿{sin 𝑎 𝑥} = 𝑠 2 +𝑎2 1 𝑎𝑥 5) 𝐿{𝑒 } = 𝑠 2 +𝑎2 𝑎 𝑠−𝑎 Step 2 Step-by-Step Explanation a) 𝐿{4𝑥 2 − 3 cos(2𝑥) + 5𝑒 𝑥 } by the linearity property = 4𝐿{𝑥 2 } − 3𝐿{cos(2𝑥)} + 5𝐿{𝑒 𝑥 } 2! 𝑠 𝑠 𝑠 2 +22 5 = 4 ( 3) − 3 ( = 8 𝑠3 − 3𝑠 𝑠 2 +4 1 + ) + 5( 1 𝑠−1 ) 𝑠−1 b) 𝐿{4 + sin(𝑥) − 𝑒 4𝑥 } 2 by the linearity property 1 = 4𝐿{1} + 𝐿{sin(𝑥)} − 𝐿{𝑒 4𝑥 } 1 2 1 = 4( ) + ( 4 = + 𝑠 1 )− 2 𝑠 2 +1 1 1 𝑠 2(𝑠 2 +1) − 1 𝑠−4 𝑠−4 Result: a) 𝐿{4𝑥 2 − 3 cos(2𝑥) + 5𝑒 𝑥 } = 1 4 2 𝑠 b) 𝐿{4 + sin(𝑥) − 𝑒 4𝑥 } = + 8 − 𝑠3 1 3𝑠 𝑠 2 +4 2(𝑠 2 +1) − + 1 5 𝑠−1 𝑠−4 Question #20 Use Laplace transforms to solve the initial value problems. 𝑦 ′′ + 4𝑦 = 2𝛿(𝑡 − 𝜋), 𝑦(0) = 1, 𝑦 ′ (0) = 0 Answer: We know that 𝐿{𝛿(𝑡 − 𝑡0 )} = 𝑒 −𝑠𝑡0 and 𝐿−1 {𝑒 −𝑎𝑠 𝐹(𝑠)} = 𝑓(𝑡 − 𝑎)𝑢(𝑡 − 𝑎) Now , given that, 𝑦 ′′ + 4𝑦 = 2𝛿(𝑡 − 𝜋); 𝑦(0) = 1. 𝑦 ′ (0) = 0 Taking Laplace transformation on both sides we get, 𝐿{𝑦 ′′ + 4𝑦} = 2𝐿{𝛿(𝑡 − 𝜋)} ⇒ 𝐿{𝑦 ′′ } + 4𝐿{𝑦} = 2𝐿{𝛿(𝑡 − 𝜋)} ⇒ 𝑠𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) + 4𝑦(𝑠) = 2𝑒 −𝜋𝑠 ⇒ 𝑠 2 𝑌(𝑠) − 𝑠 − 0 + 4𝑦(𝑠) = 2𝑒 −𝜋𝑠 [∵ 𝑦(𝑠) = 𝐿{𝑦}] ⇒ 𝑦(𝑠)[𝑠 2 + 4] = 2𝑒 −𝜋𝑠 + 𝑠 ⇒ 𝑦(𝑠) = 2 𝑒 −𝜋𝑠 𝑠 2 +4 𝑠 + 𝑠 2 +4 Now taking inverse Laplace transformation, 2𝑒 −𝜋𝑠 𝑦(𝑑) = 𝐿−1 { } + 𝐿−1 { 𝑠 2 +4 −1 −𝜋𝑠 ∴ 𝑦(𝑑) = 𝐿 {𝑒 −1 Now, 𝐿 {𝑒 2 −𝜋𝑠 Let, 𝐹(𝑠) = 2 2 𝑠 2 +22 𝑠 2 +22 𝑠 2 +22 𝑠 } 𝑠 2 +4 −1 }+𝐿 { 𝑠 𝑠 2 +22 }(2) } , Then 𝑓(𝑡) = sin 2 𝑡 ∴ 𝑓(𝑡 − 𝜋) = sin 2 (𝑡 − 𝜋) 2 𝐿−1 {𝑒 −𝜋𝑠 2 2} = 𝑓(𝑡 − 2)𝑢(𝑡 − 𝑎) 𝑠 +2 = sin(2𝑡 − 2𝜋) 𝑢(𝑡 − 𝜋) = sin 2 (𝑡 − 𝜋)𝑢(𝑡 − 𝜋) 𝑠 and 𝐿−1 { 2 2} = cos 2 𝑡 𝑠 +2 From (2) we get 𝑦(𝑡) = sin 2 (𝑡 − 𝜋)𝑢(𝑡 − 𝜋) + cos 2 𝑡 ⇒ 𝑦(𝑡) = sin 2 𝑡𝑢(𝑡 − 𝜋) + cos 2 𝑡