Questions and Answers #4 Laplace Transform

Questions and Answers Sheet 4 Laplace Transform Question #1 Solve the initial value problem using Laplace transforms. y''+y=f(t), y(0)=0, y'(0)=1 Here 0 0  t  3 f (t ) =  1 t  3 Answer: Given differential equation y''+y=f(t), y(0)=0 , y'(0)=1 Here 0 0  t  3 f (t ) =  1 t  3 To find: Solve the initial value problem using Laplace Transform We begin with taking Laplace transformation on both sides, L(y''+y)=L(f(t)) L(y'')+L(y)=L(f(t)) Note: 𝐿(𝑦 ′′ ) = 𝑠 2 𝐿(𝑦) − 𝑠𝑦(0) − 𝑦 ′ (0) Substitute in the equation, (𝑠 2 𝐿(𝑦) − 𝑠𝑦(0) − 𝑦 ′ (0)) + 𝐿(𝑦) = 𝐿(𝑓(𝑡)) We begin with finding the Laplace for RHS, 3𝜋 ∞ 𝐿(𝑓(𝑡)) = ∫0 𝑒 −𝑠𝑡 (0)𝑑𝑡 + ∫3𝜋 𝑒 −𝑠𝑡 (1)𝑑𝑡 ∞ = ∫3𝜋 𝑒 −𝑠𝑡 (1)𝑑𝑡 = = 𝑒 −𝑠𝑡 ∞ | −𝑠 3𝜋 𝑒 −3𝜋𝑠 𝑠 Now, use the conditions given, 𝑠 2 𝐿(𝑦) − 1 + 𝐿(𝑦) = 𝐿(𝑓(𝑡)) (𝑠 2 + 1)𝐿(𝑦) − 1 = (𝑠 2 + 1)𝐿(𝑦) = 𝐿(𝑦) = 𝑒 −3𝜋𝑠 𝑠(𝑠 2 +1) 𝑒 −3𝜋𝑠 𝑒 −3𝜋𝑠 + 𝑠 1 𝑠 +1 𝑠 2 +1 Now , find inverse Laplace transform 𝑒 −3𝜋𝑠 1 𝑦(𝑡) = 𝐿−1 ( 2 ) + 𝐿−1 ( 2 ) 𝑠(𝑠 + 1) 𝑠 +1 Note: 𝐿−1 {𝐹(𝑠)} = 𝑓(𝑡) then 𝐿−1 {𝑒 −𝑎𝑠 𝐹(𝑠)} = 𝐻(𝑡 − 𝑎)𝑓(𝑡 − 𝑎). Where H is Heaviside step function = 𝐻(𝑡 − 3𝜋)(𝐻(𝑡 − 3𝜋) − cos(𝑡 − 3𝜋)) + 𝑠𝑖𝑛. Therefore, 𝑦(𝑡) = 𝐻(𝑡 − 3𝜋)(𝐻(𝑡 − 3𝜋) − cos(𝑡 − 3𝜋)) + 𝑠𝑖𝑛 Question #2 Find inverse laplace transform. 1 𝐹(𝑠) = 2 𝑠(𝑠 +𝑠+1) Answer: Given: 𝐹(𝑠) = Now, 1 𝑠(𝑠 2 +𝑠+1) 1 𝐴 𝐵𝑠+𝐶 𝑠 𝑠 2 +𝑠+1 = + 𝑠(𝑠 2 +𝑠+1) = 𝐴(𝑠 2 +𝑠+1)+(𝐵𝑠+𝐶)𝑠 𝑠(𝑠 2 +𝑠+1) Compairing with both sicly we get 1 = 𝐴𝑠 2 + 𝐴𝑠 + 𝐴 + 𝐵𝑠 2 + 𝐶𝑠 = (𝐴 + 𝐵)𝑠 2 + (𝐴 + 𝐶)𝑠 + 𝐴 Therefore. A=1 A+B=0 A+C=0 1+B=0 1+C=0 B=-1 C=-1 Therefore 1 1 𝑠+1 = − 2 2 𝑠(𝑠 +𝑠+1) 1 𝑠 𝐹(𝑠) = − 𝑠 𝑠 +𝑠+1 𝑠+1 1 2 1 𝑠+1 𝑠 (𝑠+ ) + = − 1 (𝑠+ ) +( ) 2 = − 2 1 𝑠+2 2 1 2 √3 (𝑠+ ) +( ) 2 2 1 𝑠+2 1 𝑠 1 4 1 2 3 2 4 1 1 𝑠+2+2 2 1 2 √3 = − 𝑠 1 4 𝑠 2 +2 𝑠+ +1− − 1 = − 𝑠 1 2 2 (𝑠+ ) +( √3 ) 2 − 2 1 1 2 2 1 2 √3 (𝑠+ ) +( ) 2 2 √3 2 2 √3 1 2 √3 (𝑠+ ) +( ) 2 2 1 the inverse of F(s) 1 𝑠+2 1 𝐿−1 {𝐹(𝑠)} = 𝐿−1 { } − 𝐿−1 { 2 1 2 √3 (𝑠+ ) +( ) 2 2 𝑡 √3 − 2 𝑠 𝑡 = 1 − 𝑒 −2 cos ( √3 2 𝑡) − 1 √3 𝑒 sin ( }− 1 √3 𝐿−1 { √3 2 2 2 1 √3 (𝑠+ ) +( ) 2 2 } 𝑡) 2 Question #3 Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. 𝑦 ′′ + 5𝑦 = 2𝑡 2 − 9, 𝑦(0) = 0, 𝑦 ′ (0) = −3 Answer: 𝑦 ′′ + 5𝑦 = 2𝑡 2 − 9 𝑦(0) = 0, 𝑦 ′ (0) = −3 𝐿{𝑦 ′′ } + 𝑠𝐿{𝑦 ′ } = 𝐿{2𝑡 2 − 9} 𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) + 5𝑌(𝑠) = 𝐿{2𝑡 2 } − 𝐿{9} 4 9 = 3− 𝑠 𝑠 2 ⇒ 𝑠 𝑌(𝑠) − 𝑠𝑦(0) + 3 + 5𝑌(𝑠) = ⇒ 𝑌(𝑠)[𝑠 2 + 5] = ⇒ 𝑌(𝑠) = ⇒ 𝑦(𝑡) = = 2𝑡 2 − 5 − 𝑠 𝑠 − 9 𝑠 3 − −3 4−9𝑠 2 −3𝑠 3 𝑠 3 (𝑠 2 +5) 2 3 −1 4−9𝑠 −3𝑠 𝐿 { 3 (𝑠2 } 𝑠 +5) 3√5 sin(√5𝑡) 25 9 = 5 4−9𝑠 2 −3𝑠 3 [∵ 𝐿−1 { 3(𝑠2 } 𝑠 +5) 2 2𝑡 49 3√5 ∴ 5 4 9 − −3 𝑠3 𝑠 𝑠 2 +5 4 4 𝑠3 − 25 + = 49 cos(√5𝑡) 25 4 5𝑠 3 − − 15 𝑠(𝑠 2 +25) 49 sin(√5𝑡) + 25 49 25 + 49𝑠 25(25+52 ) cos(√5𝑡) − 49 25𝑠 2 ] Question #4 Solve IVPs by Laplace Transform 𝑦 ′′ + 3𝑦 ′ − 4𝑦 = 6𝑒 2𝑡−3 y(1.5)=4 y'(1.5)=5 Answer: Step 1 Given y′′ + 3𝑦 ′ − 4𝑦 = 6𝑒 2𝑡−3 y(1.5)=4 y'(1.5)=5 Since initial values are not given at t=0 , so we need to change the variable Let 𝑡 = 𝑧 + 1.5 ⇒ 𝑧 = 𝑡 − 1.5 So, our original equation is 𝑦 ′′ (𝑡) + 3𝑦 ′ (𝑡) − 4𝑦(𝑡) = 6𝑒 2𝑡−3 which now becomes 𝑦 ′′ (𝑧 + 1.5) + 3𝑦 ′ (𝑧 + 1.5) − 4𝑦(𝑧 + 1.5) = 6𝑒 2(𝑧+1.5)−3 = 𝑦 ′′ (𝑧 + 1.5) + 3𝑦 ′ (𝑧 + 1.5) − 4𝑦(𝑧 + 1.5) = 6𝑒 2𝑧 Let 𝑢(𝑧) = 𝑦(𝑧 + 1.5) 𝑑𝑢 𝑑𝑦 𝑑𝑡 ⇒ 𝑢′ (𝑧) = = = 𝑦 ′ (𝑧 + 1.5) 𝑑𝑧 𝑑𝑡 𝑑𝑧 similarly 𝑢′′ (𝑧) = 𝑦 ′′ (𝑧 + 1.5) The initial conditions for u(z) are u(0)=y(1.5)=4 𝑢′ (0) = 𝑦 ′ (1.5) = 5 Step 2 So, the differential equation is 𝑢′′ (𝑧) + 3𝑢′ (𝑧) − 4𝑢(𝑧) = 6𝑒 2𝑧 with u(0)=4, u'(0)=5 Taking Laplace transform both sides, we get 6 𝑠 2 𝑢̅ − 𝑠𝑢(0) − 𝑢′ (0) + 3𝑠𝑢̅ − 3𝑢(0) − 4𝑢̅ = ⇒ 𝑠 2 𝑢̅ − 4𝑠 − 5 + 3𝑠𝑢̅ − 12 − 4𝑢̅ = ⇒ (𝑠 2 + 3𝑠 − 4)𝑢̅ = ⇒ (𝑠 2 + 3𝑠 − 4)𝑢̅ = 6 6 𝑠−2 𝑠−2 + 4𝑠 + 17 𝑠−2 4𝑠 2 +9𝑠−28 𝑠−2 4𝑠 2 +9𝑠−28 ⇒ 𝑢̅ = (𝑠−2)(𝑠2 +3𝑠−4) (4𝑠−7)(𝑠+4) ⇒ 𝑢̅ = (𝑠−2)(𝑠+4)(𝑠−1) (4𝑠−7) ⇒ 𝑢̅ = (𝑠−2)(𝑠−1) Step 3 Now, by partial fraction, (4𝑠−7) (𝑠−2)(𝑠−1) = 𝐴 𝑠−2 + 𝐵 𝑠−1 ⇒ 4𝑠 − 7 = 𝐴(𝑠 − 1) + 𝐵(𝑠 − 2) when s=1 B=3 when s=2 A=1 so, (4𝑠−7) = 1 (𝑠−2)(𝑠−1) 𝑠−2 (4𝑠−7) + so, 𝑢̅ = (𝑠−2)(𝑠−1) = 3 𝑠−1 1 𝑠−2 + 3 𝑠−1 On taking Inverse Laplace transformation, we get 𝑢(𝑧) = 𝑒 2𝑧 + 3𝑒 −𝑧 Now, 𝑦(𝑡) = 𝑦(𝑧 + 1.5) = 𝑢(𝑧) = 𝑢(𝑡 − 1.5) so, 𝑢(𝑡 − 1.5) = 𝑒 2(𝑡−1.5) + 3𝑒 −(𝑡−1.5) ⇒ 𝑦(𝑡) = 𝑒 2(𝑡−1.5) + 3𝑒 −(𝑡−1.5) is the answer Question #5 Find the Laplace Transform of the function. 𝑓(𝑥) = sin ℎ (𝑎𝑥) Answer: 𝑓(𝑥) = sin ℎ (𝑎𝑥) ⇒ 𝑓(𝑥) = 𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2 𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 ; sin ℎ (𝑎𝑥) = 2 Taking Laplace transform on both sides , we get 𝐿{𝑓(𝑥)} = 𝐿{ 𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2 −𝑎𝑥 1 = [𝐿{𝑒 𝑎𝑥 } − 𝐿{𝑒 2 We have 𝐿{𝑒 𝑎𝑥 } = 1 } }] : using linearity property ; 𝑠 > 𝑎 , then 𝑠−𝑎 1 1 𝐿{𝑓(𝑥)} = [ 1 − 2 𝑠−𝑎 𝑠−(−𝑎) 1 1 1 ⇒ 𝐿{𝑓(𝑥)} = [ 2 𝑠−𝑎 1 (𝑠+𝑎)−(𝑠−𝑎) − ] ; 𝑠 > 𝑎, 𝑠 ≻ 𝑎 𝑠+𝑎 ] = [ (𝑠−𝑎)(𝑠+𝑎) ] 2 1 𝑠+𝑎−𝑠+𝑎 = [ 2 𝑠 2 −𝑎2 ] ; ∵ (𝑏 − 𝑐)(𝑏 + 𝑐) = 𝑏 2 − 𝑐 2 1 2𝑎 2 𝑠 2 −𝑎2 ⇒ 𝐿{𝑓(𝑥)} = × ⇒ 𝐿{𝑓(𝑥)} = 𝑎 𝑠 2 −𝑎2 ; 𝑠 > 𝑎, 𝑠 ≻ 𝑎 or 𝑠 > |𝑎| Hence , we get 𝐿{𝑓(𝑥)} = 𝐿{sin ℎ (𝑎𝑥)} = 𝑎 𝑠 2 −𝑎2 ; 𝑠 > |𝑎| Question #6 Find the laplace transform of 11𝑠+28 𝐹(𝑠) = (𝑠+2)2(𝑠+5) ⇒ 𝑓(𝑡) = 𝐿−1 [𝐹(𝑠)] =? Answer: Solution: 11𝑠+28 Given: 𝐹(𝑠) = (𝑠+2)2(𝑠+5) Find 𝑓(𝑡) = 𝐿−1 [𝐹(𝑠)] i.e. Find the inverse Laplace Transform of 11𝑠+28 (𝑠+2)2 (𝑠+5) use the partial fraction 11𝑠+28 𝐴 𝐵 𝐶 = (𝑠+2) + (𝑠+2)2 + (𝑠+5) (𝑠+2)2 (𝑠+5) 11𝑠+28 ⇒ (𝑠+2)2(𝑠+5) = 𝐴(𝑠+2)(𝑠+5)+𝐵(𝑠+5)+𝐶(𝑠+2)2 (𝑠+2)2 (𝑠+5) ⇒ 11𝑠 + 28 = 𝐴(𝑠 + 2)(𝑠 + 5) + 𝐵(𝑠 + 5) + 𝐶(𝑠 + 2)2 At 𝑠 = −2,11 × −2 + 28 = 𝐴(0)(3) + 𝐵(3) + 𝐶 ⋅ 02 ⇒ 28 − 22 = 0 + 3𝐵 + 0 ⇒ 3𝐵 = 6 ⇒𝐵=2 ∴𝐵=2 At 𝑠 = −5,11 × −5 + 28 = 𝐴(−3)0 + 𝐵 ⋅ 0 + 𝐶(−3)2 ⇒ 28 − 55 = 0 + 0 + 9𝐶 ⇒ 9𝐶 = −27 ⇒ 𝐶 = −3 ∴ 𝐶 = −3 At s=0, 11 × 0 + 28 = 𝐴(2)(5) + 𝐵 ⋅ 5 + 𝐶 ⋅ 22 ⇒ 28 = 10𝐴 + 5𝐵 + 4𝐶 ⇒ 28 = 10𝐴 + 5 × 2 + 4 × −3[∵ 𝐵 = 2, 𝐶 = −3] ⇒ 28 = 10𝐴 + 10 − 12 = 10𝐴 − 2 ⇒ 10𝐴 = 28 + 2 ⇒ 10𝐴 = 30 ⇒ 𝐴 = 3 ∴𝐴=3 11𝑠 + 28 3 2 −3 ∴ = + + (𝑠 + 2)2 (𝑠 + 5) (𝑠 + 2) (𝑠 + 2)2 (𝑠 + 5) 3 2 3 So, 𝑓(𝑡) = 𝐿−1 [(𝑠+2) + (𝑠+2)2 − (𝑠+5)] 1 1 1 = 3𝐿−1 [(𝑠+2)] + 2𝐿−1 [(𝑠+2)2] − 3𝐿−1 [(𝑠+5)] = 3𝑒 −2𝑡 + 2𝑡𝑒 −2𝑡 − 3𝑒 −5𝑡 Thus 𝑓(𝑡) = 3𝑒 −2𝑡 + 2𝑡𝑒 −2𝑡 − 3𝑒 −5𝑡 Question #7 Find the inverse Laplace transform of the following transfer function: 𝑌(𝑠) 𝑈(𝑠) 50 = (𝑠+7)2 +25 Select one: a) 𝑓(𝑡) = 10𝑒 −7𝑡 sin(5𝑡) b) 𝑓(𝑡) = 10𝑒 7𝑡 sin(5𝑡) c) 𝑓(𝑡) = 50𝑒 −7𝑡 sin(5𝑡) d) 𝑓(𝑡) = 2𝑒 −7𝑡 sin(5𝑡) Answer: We know that 𝑌(𝑠) 𝑈(𝑠) 50 = (𝑠+7)2 +25 𝑎 𝐿{sin 𝑎 𝑡} = 2 𝑠 + 𝑎2 also 𝐿{𝑒 𝑎𝑡 𝑓(𝑡)} = 𝐹(𝑠 − 𝑎) Where 𝐿{𝑓(𝑡)} = 𝐹(𝑠) 𝑌(𝑠) 50 10 ⋅ 5 = = 𝑈(𝑠) (𝑠 + 7)2 + 25 (𝑠 + 7)2 + 52 so , 𝐿−1 { 𝑌(𝑠) 𝑈(𝑠) 10⋅5 } = 𝐿−1 {(𝑠+7)2 +52 } −7𝑡 = 10𝑒 sin(5𝑡) So, 𝑓(𝑡) = 10𝑒 −7𝑡 sin(5𝑡) Thus a) 𝑓(𝑡) = 10𝑒 −7𝑡 sin(5𝑡) is correct Question #8 3 Find the laplace transform of 𝐿{𝑓(𝑥)} being 𝑓(𝑥) = ∫0 cos(3𝑡) 𝑑𝑡 Answer: 3 Find 𝐿{𝑓(𝑥)} where 𝑓(𝑥) = ∫0 cos(3𝑡) 𝑑𝑡 3 Given that 𝑓(𝑥) = ∫0 cos(3𝑡) 𝑑𝑡, so we first calculate this integral. 3 3 𝑓(𝑥) = ∫ cos(3𝑡) 𝑑𝑡 = [ 0 ⇒ 𝑓(𝑥) = sin(3𝑡) 1 ] = [sin 9 − sin 0] 3 3 0 sin 9 3 ⇒ 𝑓(𝑥) is a constant function ∞ Now 𝐹(𝑠) = 𝐿{𝑓(𝑥)} = ∫0 𝑓(𝑥)𝑒 −𝑠𝑥 dx ∞ sin 9 = ∫0 3 𝑒 −𝑠𝑥 𝑑𝑥 = sin 9 3 ∞ ∫0 𝑒 −𝑠𝑥 𝑑𝑥 = sin 9 𝑒 −𝑠𝑥 3 [ −𝑠 ∞ ] = − sin 9 3𝑠 0 ⇒ 𝐹(𝑠) = 𝐿{𝑓(𝑥)} = [𝑒 −∞ − 𝑒 0 ] 1 3𝑠 sin 9 Question #9 Solve the initial value problem below using the method of Laplace transforms. 𝑦 ′′ − 8𝑦 ′ + 41𝑦 = 29𝑒 2𝑡 y(0)=1 , y'(0)=7 Answer: The given IVP is 𝑦 ′′ − 8𝑦 + 41𝑦 = 29𝑒 2𝑡 y(0)=1 , y'(0)=7 Taking Laplace in both side we get 𝐿[𝑦 ′′ ] − 8𝐿[𝑦 ′ ] + 41𝐿[𝑦] = 29𝐿[𝑒 2𝑡 ] ⇒ 𝑠 2 𝑦(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) − 8(𝑠𝑦(𝑠) − 𝑦(0)) + 41𝑦(𝑠) = 29 29 𝑠−2 ⇒ (𝑠 2 − 8𝑠 + 41)𝑦(𝑠) − 𝑠(1) − 7 − 8(−1) = (𝑠−2) 29 ⇒ (𝑠 2 − 8𝑠 + 41)𝑦(𝑠) − 𝑠 − 7 + 8 = (𝑠−2) 29 ⇒ (𝑠 2 − 8𝑠 + 41)𝑦(𝑠) − 𝑠 + 1 = (𝑠−2) 29 ⇒ (𝑠 2 − 8𝑠 + 41)𝑦(𝑠) = (𝑠−2) + 𝑠 − 1 ⇒ (𝑠 2 − 8𝑠 + 41)𝑦(𝑠) = 29+𝑠 2 −3𝑠+2 (𝑠−2) 𝑠 2 − 3𝑠 + 31 ⇒ 𝑦(𝑠) = (𝑠 − 2)(𝑠 2 − 8𝑠 + 41) ⇒ 𝑦(𝑠) = 𝑠 2 −3𝑠−5𝑠+5𝑠+31+10−10 (𝑠−2)(𝑠 2 −8𝑠+41) 𝑠 2 − 8𝑠 + 5𝑠 + 41 − 10 ⇒ 𝑦(𝑠) = (𝑠 − 2)(𝑠 2 − 8𝑠 + 41) ⇒ 𝑦(𝑠) = (𝑠 2 −8𝑠+41)+5(𝑠−2) (𝑠−2)(𝑠 2 −8𝑠+41) (𝑠 2 −8𝑠+41) or 𝑦(𝑠) = (𝑠−2)(𝑠2 −8𝑠+41) 5(𝑠−2) + (𝑠−2)(𝑠2 −8𝑠+41) 1 5 𝑦(𝑠) = + (𝑠 − 2) (𝑠 2 − 8𝑠 + 41) 1 5 or 𝑦(𝑠) = (𝑠−2) + (𝑠2 −8𝑠+16+25) 5 1 ⇒ 𝑦(𝑠) = (𝑠−2) + (𝑠−4)2 +52 (1) 𝑎𝑡 Since we know 𝐿[𝑒 𝑓(𝑡)] = 𝐹(𝑠 − 𝑎) where 𝐹(𝑠) = 𝐿[𝑓(𝑡)] i.e. we just have to put (s-a) at place of s in F(s) (Laplace of f(t)) to find Laplace of 𝑒 𝑎𝑡 ⋅ 𝑓(𝑡) ∵ for 𝑓(𝑡) = sin(5𝑡) ⇒ 𝐹(𝑠) = 5 𝑠 2 +52 So for finding Laplace of 𝑒 4𝑡 sin(5𝑡) we just have to put s-4 at the place of s in F(s) Then 5 𝐿[𝑒 4𝑡 sin(5𝑡)] = (𝑠−4)2 +52 5 ] = 𝑒 4𝑡 sin(5𝑡) (𝑠 − 4)2 + 52 Now Taking inverse Laplace of (1) we get ⇒ 𝐿−1 [ 𝑦(𝑡) = 𝐿−1 [ 1 𝑠−2 2𝑡 5 ] + 𝐿−1 [(𝑠−4)2 4𝑡 +52 ] ⇒ 𝑦(𝑡) = 𝑒 + 𝑒 sin(5𝑡) This is required solution of given IVP Question #10 1 If 𝑋(𝑠) = (𝑠−1)(𝑠+1), find x(t) Answer: Step 1 Let us find x(t) Given: 1 𝑋(𝑠) = (𝑠−1)(𝑠+1) Inverse Laplace transform: 1 1 𝐿−1 { 2 2 } = sin ℎ (𝑎𝑡) 𝑠 −𝑎 𝑎 Step 2 Simplify the given X(s) 1 𝐿(𝑋(𝑠)) = (𝑠−1)(𝑠+1) (𝑠 − 1)(𝑠 + 1) = 𝑠 2 − 12 𝐿(𝑋(𝑠)) = 1 𝑠 2 −12 𝐿(𝑋(𝑠)) = 𝑋(𝑡) 𝑋(𝑡) = 𝐿−1 ( 1 𝑠 2 −12 ) a=1 1 𝑋(𝑡) = sin ℎ (1𝑡) 1 𝑋(𝑡) = sin ℎ 𝑡 Step 3 Answer 𝑋(𝑡) = sin ℎ 𝑡 Question #11 Find 𝐿 [ 𝑒 −3𝑡 −cos 2𝑡 𝑡 ] Answer: Step 1 We need to find the Laplace transformation, 𝑒 −3𝑡 −cos 2𝑡 𝐿[ 𝑡 ] Note: ∞ 𝑓(𝑡) 𝐿{ 𝑡 ∞ } = ∫𝑠 𝐿{𝑓(𝑡)}𝑑𝑢 = ∫𝑠 𝐹(𝑢)𝑑𝑢 Here, f(t)=e^{-3t}-\cos 2t Step 2 𝐿{𝑓(𝑡)} = 𝐿{𝑒 −3𝑡 − cos(2𝑡)} = ( 𝑒 −3𝑡 −cos 2𝑡 Now, 𝐿{ = 𝑡 ∞ } = ∫𝑠 ( − ∞ 1 ∫𝑠 𝑢+3 𝑑𝑢 = ln(𝑢 + = ln ( ∞ 𝑢 − ∫𝑠 2 𝑑𝑢 𝑢 +4 1 3) |∞ − ln(𝑢2 + 𝑠 2 1 𝑢+3 𝑢+3 √𝑢2 +4 = ln ( ) |∞ 𝑠 3 𝑢 4 √1+ 2 𝑢 1+ ) |∞ 𝑠 4) |∞ 𝑠 1 𝑠+3 𝑢 − 𝑢2 +4 𝑠 𝑠 2 +4 ) 𝑑𝑢 ) = ln 1 − ln ( 3 𝑠 4 √1+ 2 𝑠 1+ 𝑠+3 = 0 − ln ( √𝑠 2 +4 𝑠+3 = − ln ( √𝑠 2 +4 = ln ( √𝑠 2 +4 𝑠+3 ) ) ) ) Question #12 Show that 𝐿{sin 𝑘 𝑡} = 𝐿{cos 𝑘 𝑡} = 𝑠 𝑠 2 +𝑘 2 𝑘 𝑠 2 +𝑘 2 ∞ Use 𝐿{sin 𝑘 𝑡} = ∫0 𝑒 −𝑠𝑡 sin 𝑘 𝑡𝑑𝑡 1) 𝐿{1} = 2) 𝐿{𝑡 𝑛 } = 3) 𝐿{𝑒 𝑘𝑡 } = 4) 𝐿{sin 𝑘 𝑡} = 5) 𝐿{cos 𝑘 𝑡} = 6) 𝐿{sin ℎ 𝑘𝑡} = 7) 𝐿{cos ℎ 𝑘𝑡} = Answer: Step 1 We have to laplace transform of sin kt and c t Step 2 Show that 𝐿{sin 𝑘 𝑡} = 𝐿{cos 𝑘 𝑡} = 𝑠 𝑘 𝑠 2 +𝑘 2 𝑠 2 +𝑘 2 Solution: We know that ∞ 𝐿{\𝑓(𝑡)} = ∫0 𝑒 −𝑠𝑡 𝑓(𝑡)𝑑𝑡 ∞ Now, 𝐿{sin 𝑘 𝑡} = ∫0 𝑒 −𝑠𝑡 sin 𝑘 𝑡𝑑𝑡 ∞ = 𝐿{sin 𝑘 𝑡} = ∫0 𝑒 −𝑠𝑡 sin 𝑘 𝑡𝑑𝑡 𝑒 −𝑠𝑡 = [((−𝑠)2 +𝑘 2 ) 0 𝑒0 = 0 − (𝑠2 +𝑘 2 ) (−0 − 𝑘) = Hence 𝐿{sin 𝑘 𝑡} = Again 𝐿{cos 𝑘 𝑡} = 𝑒 −𝑠𝑡 = [((−𝑠)2 =0−( +𝑘 2 ) ∞ (−𝑠 sin 𝑘 𝑡 − 𝑘 cos 𝑘 𝑡)] 𝑘 𝑠 2 +𝑘 2 𝑘 𝑠 2 +𝑘 2 ∞ ∫0 𝑒 −𝑠𝑡 cos 𝑘 𝑡𝑑𝑡 ∞ (−𝑠 cos 𝑘 𝑡 + 𝑘 sin 𝑘 𝑡)] 𝑒0 𝑠 2 +𝑘 2 0 (−𝑠 + 0)) = Hence , 𝐿{cos 𝑘 𝑡} = (𝑠2 𝑠 𝑠 2 +𝑘 2 𝑠 +𝑘 2 ) Note: (1) ∫ 𝑒 𝑎𝑥 sin 𝑏 𝑥𝑑𝑥 = (2) ∫ 𝑒 𝑎𝑥 cos 𝑏 𝑥𝑑𝑥 = (3) 𝐿{1} = 1 3 (4) 𝐿{𝑡 𝑛 } = 𝑛! 𝑠 𝑛+1 𝑒 𝑎𝑥 𝑎2 +𝑏 2 𝑒 𝑎𝑥 𝑎2 +𝑏 2 (𝑎 sin 𝑏 𝑥 − 𝑏 cos 𝑏 𝑥) (𝑎 cos 𝑏 𝑥 + 𝑏 sin 𝑏 𝑥) (5) 𝐿{𝑒 𝑘𝑡 } = 1 𝑠−𝑘 (6) 𝐿{sin 𝑘 𝑡} = (7) 𝐿{cos 𝑘 𝑡} = 𝑘 𝑠 2 +𝑘 2 𝑠 𝑠 2 +𝑘 2 𝑘 (8) 𝐿{sin ℎ 𝑘𝑡} = (9) 𝐿{cos ℎ 𝑘𝑡} = 𝑠 2 −𝑘 2 𝑠 𝑠 2 −𝑘 2 Question #13 𝑡 Find Laplace transform 𝐿{10𝑒 −5𝑡 sin } 2 Answer: Solution: 𝑡 𝐿{10𝑒 −5𝑡 sin ( )} 2 𝑡 = 10𝐿{𝑒 −5𝑡 sin ( )} 2 as we know that 𝑡 2 2 4(𝑠+5)2 +1 𝑡 𝑒 −5𝑡 sin ( ) ⇒ ∴ 𝐿{10𝑒 −5𝑡 sin ( )} ⇒ 2 20 4(𝑠+5)2 +1 Question #14 Find 𝐿{9 + 6𝑡} 𝐿{𝑘} = 𝐿{𝑡} = 𝑘 𝑠 1 𝑠2 Answer: Step 1 Given The given expression is 𝐿{9 + 6𝑡} Apply Laplace transform formula 𝑘 𝐿{𝑘} = , where k = constant 𝐿{𝑡} = 𝑠 1 𝑠2 Step 2 Calculation 𝐿{9 + 6𝑡} = 𝐿{9} + 𝐿{6𝑡} 9 6 = + 2 𝑠 𝑠 9𝑠+6 = 2 = 𝑠 3(3𝑠+2) 𝑠2 Hence, 𝐿{9 + 6𝑡} = Question #15 Find f(t) 𝐿−1 { (3𝑠−1) 𝑠 2 (𝑠+1)3 } 3(3𝑠+2) 𝑠2 Answer: Step 1 From the given information it is needed to calculate: 𝐿−1 { (3𝑠−1) 𝑠 2 (𝑠+1)3 } Step 2 Now, first take the partial fractions of the given expression and solve: (3𝑠−1) = 𝑎0 + 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 (𝑠+1) (𝑠+1)2 (𝑠+1)3 𝑠 2 (𝑠+1)3 𝑠 𝑠2 𝑎0 (𝑠)(𝑠+1)3 +𝑎1 (𝑠+1)3 +𝑎2 (𝑠 2 )(𝑠+1)2 +𝑎3 (𝑠 2 )(𝑠+1)+𝑎4 (𝑠 2 ) = 𝑠 2 (𝑠+1)3 1)3 + 𝑎1 (𝑠 3𝑠 − 1 = 𝑎0 (𝑠)(𝑠 + + 1)3 + 𝑎2 (𝑠 + 1)2 + 𝑎3 (𝑠 2 )(𝑠 + 1) + 𝑎4 (𝑠 2 ) when 𝑠 = 0 ⇒ 3(0) − 1 = 𝑎1 (1)3 ⇒ 𝑎1 = −1 when 𝑠 = −1 ⇒ 3(−1) − 1 = 𝑎4 (−1)2 ⇒ −4 = 𝑎4 Step 3 Now calculate rest of the unknowns by putting the value of the 𝑎1 and 𝑎4 : 3𝑠 − 1 = 𝑎0 (𝑠)(𝑠 + 1)3 + (−1)(𝑠 + 1)3 + 𝑎2 (𝑠 2 )(𝑠 + 1)2 + 𝑎3 (𝑠 2 )(𝑠 + 1) + (−4)(𝑠 2 ) 3𝑠 − 1 = 𝑎0 𝑠 4 + 𝑎2 𝑠 4 + 3𝑎0 𝑠 3 + 2𝑎2 𝑠 3 + 𝑎3 𝑠 3 − 𝑠 3 + 3𝑎0 𝑠 2 + 𝑎2 𝑠 2 + 𝑎3 𝑠 2 − 7𝑠 2 + 𝑎0 𝑠 − 3𝑠 − 1 3𝑠 − 1 = 𝑠 4 (𝑎0 + 𝑎2 ) + 𝑠 3 (3𝑎0 + 2𝑎2 + 𝑎3 − 1) + 𝑠 2 (3𝑎0 + 𝑎2 + 𝑎3 − 7) + 𝑠(𝑎0 − 3) − 1 now by comparing: 𝑎0 + 𝑎2 = 0 3𝑎0 + 2𝑎2 + 𝑎3 − 1 = 0 3𝑎0 + 𝑎2 + 𝑎3 − 7 = 0 𝑎0 − 3 = 3 ⇒ 𝑎0 = 6 so , 𝑎0 + 𝑎2 = 0 ⇒ 6 + 𝑎2 = 0 ⇒ 𝑎2 = −6 and , 3𝑎0 + 2𝑎2 + 𝑎3 − 1 = 0 ⇒ 3(6) + 2(−6) + 𝑎3 − 1 = 0 18 − 12 − 1 + 𝑎3 = 0 ⇒ 𝑎3 = −5 Step 4 Now, put the value of unknowns: (3𝑠 − 1) (−6) 6 −1 −5 −4 = + 2 + + + 2 3 2 (𝑠 + 1) (𝑠 + 1) (𝑠 + 1)3 𝑠 (𝑠 + 1) 𝑠 𝑠 (3𝑠−1) (−6) 6 −1 −5 −4 } = 𝐿−1 { + 2 + (𝑠+1) + (𝑠+1)2 + (𝑠+1)3} 𝑠 2 (𝑠+1)3 𝑠 𝑠 (−6) 6 −1 −5 −4 𝐿−1 { } + 𝐿−1 { 2 } + 𝐿−1 {(𝑠+1)} + 𝐿−1 {(𝑠+1)2} + 𝐿−1 {(𝑠+1)3} 𝑠 𝑠 −𝑡 ) −𝑡 −𝑡 2 ) 𝐿−1 { = = 6(1) − 𝑡 − 6(𝑒 − 5(𝑒 𝑡) − 4(2𝑒 𝑡 = 6 − 𝑡 − 6𝑒 −𝑡 − 5𝑒 −𝑡 𝑡 − 8𝑒 −𝑡 𝑡 2 Step 5 The formula used in the above inverse laplace transform are: 1 𝐿−1 ( ) = 1 𝑠 𝐿−1 ( 𝐿−1 ( 𝑛! 𝑠 𝑛+1 1 𝑠−𝑎 ) = 𝑡𝑛 ) = 𝑒 𝑎𝑡 Question #16 Determine the Inverse Laplace Transform of the functions by reduction to partial fractions. 𝑠 2 −𝑠+3 𝑠 3 (𝑠+5)(𝑠−2) Answer: Step 1 Given: 𝑠 2 −𝑠+3 𝑠 3 (𝑠+5)(𝑠−2) Step 2 𝑠 2 −𝑠+3 𝐿−1 { } 𝑠 3 (𝑠+5)(𝑠−2) On taking partial function 127 1 3 = 𝐿−1 {− + 2 − 3 + 1000𝑠 100𝑠 10𝑠 33 875(𝑠+5) + 5 56(𝑠−2) } Using the linearity property of inverse Laplace Transform 127 1 33 −1 3 −1 = 𝐿−1 {− } + 𝐿−1 { } + 𝐿−1 { 2} − 𝐿 { 3} + 𝐿 { 1000𝑠 100𝑠 10𝑠 875(𝑠+5) 5 } 56(𝑠−2) On simplifying =− 127 1000 𝐻(𝑡) + 𝑡 100 − 3𝑡 2 20 + 33 875 𝑒 −5𝑡 + 5 56 𝑒 2𝑡 Question #17 use the inverse form of the First translation theorem to find the following laplace transform 𝑠 𝐿−1 { 2 } 𝑠 +4𝑠+5 Answer: Step 1 In Laplace Transform, properties used. 1 1 𝐿−1 [ 2 ] = sin 𝑎 𝑡 2 𝑠 +𝑎 𝑎 𝑠 −1 𝐿 [ 2 ] = cos 𝑎 𝑡 𝑠 + 𝑎2 Shifting property 𝐿−1 [𝑓(𝑠 − 𝑎)] = 𝑒 𝑎𝑡 𝑓(𝑡) Step 2 Consider 𝑠 𝑠 2 +4𝑠+5 Use competition of squares for denominator 𝑠 2 + 4𝑠 + 5 = 𝑠 2 + 2 × 2𝑠 + 22 − 22 + 5 = (𝑠 + 2)2 + 12 Step 3 𝑠 𝑠 𝑠 2 +4𝑠+5 = (𝑠+2)2 𝑠+2−2 +12 = (𝑠+2)2 +12 spliting the equation (𝑠+2) 2 (𝑠+2)2 +12 − (𝑠+2)2 +12 Taking Laplace inverse for the equation. (𝑠+2) 𝐿−1 [(𝑠+2)2 ⇒ 𝑒 −2𝑡 𝐿−1 [ −2𝑡 2 +12 ] − 𝐿−1 [(𝑠+2)2 𝑠 𝑠 2 +12 +12 ] ] − 𝑒 −2𝑡 2𝐿−1 [ 1 𝑠 2 +12 −2𝑡 =𝑒 cos 𝑡 − 2𝑒 𝑠𝑖𝑛. = 𝑒 −2𝑡 (cos 𝑡 − 2 sin 𝑡) Question #18 Inverse Laplace transform 2𝑠 𝐿−1 { } (𝑠 − 3)2 ] Answer: Step 1 2𝑠 Here, 𝑓(𝑠) = (𝑠−3)2 Step 2 Take the partial fraction of f(s) 2𝑠 𝐿−1 {(𝑠−3)2} 2𝑠 𝑎 = (𝑠−3)2 𝑠−3 𝑎(𝑠−3)+𝑏 = = 𝑏 + (𝑠−3)2 (𝑠−3)2 𝑎𝑠−3𝑎+𝑏 (𝑠−3)2 2𝑠 = 𝑎𝑠 − 3𝑎 + 𝑏 Step 3 Comparing coefficients of s a=2 Comparing constant terms. -3a+b=0 -3(2)+b=0 b=6 So, function will be, 2 6 𝑓(𝑠) = + 𝑠 − 3 (𝑠 − 3)2 𝐿−1 𝑓(𝑠) = 𝐿−1 ( 2 6 𝑠−3 ) + 𝐿−1 ((𝑠−3)2) Step 4 Solving first part. 𝐿−1 ( = 2𝑒 2 𝑠−3 3𝑡 ) = 2𝐿−1 ( 1 𝑠−3 ) Step 5 Solving second part. 6 6 𝐿−1 ((𝑠−3)2) = 𝑒 3𝑡 𝐿−1 { 2 } 𝑠 = 6𝑒 3𝑡 𝑡 2−1 = 6𝑡𝑒 3𝑡 Step 6 Determine inverse laplace of the function. 𝐿−1 {𝑓(𝑠)} = 2𝑒 3𝑡 + 6𝑡𝑒 3𝑡 Question #19 Find inverse Laplace transform 10 𝐹(𝑠) = 3 2 𝑠 +4𝑠 +9𝑠+10 Answer: 𝐹(𝑠) = 10 𝑠 3 +4𝑠 2 +9𝑠+10 Now using partial fraction of 10 𝑠 3 +4𝑠 2 +9𝑠+10 2 𝑠 3 + 4𝑠 2 + 9𝑠 + 10 = (𝑠 + 2)(𝑠 + 2𝑠 + 5) 10 = 𝑎0 (𝑠−2)(𝑠 2 +2𝑠+5) 10 = 𝑎0 (𝑠 2 + 10 = 𝑠 2 (𝑎2 + 𝑠+2 (𝑠+2)(𝑠 +2𝑠+5) 𝑠+2 + 𝑎2 𝑠+𝑎1 𝑠 2 +2𝑠+5 2𝑠 + 5) + (𝑎2 𝑠 + 𝑎1 )(𝑠 + 2) 2) + 𝑠(𝑎1 + 2𝑎2 + 4) + (2𝑎1 + 10) ⋯ (3) Now equate coefficients 10 = 10 + 2𝑎1 , 4 + 2𝑎2 + 𝑎1 = 0 2 + 𝑎2 = 0 𝑎2 = −2, 𝑎1 = 0, 𝑎0 = 2(4) from 2 ,4 10 2 2𝑠 (5) = − 2 2 𝑠 +2𝑠+5 taking inverse laplace both sides we get 10 2 𝑠 𝐿−1 {(𝑠+2)(𝑠2 } = 𝐿−1 { } − 2𝐿−1 { 2 +2𝑠+5) from (6) 𝐿1 = 𝐿−1 { 2 𝑠+2 𝐿2 = 𝐿−1 { 𝑠 +2𝑠+5 𝑠 }= = 𝐿 {(𝑠+1)2 } +4 (𝑠+1) 1 = 𝐿−1 {(𝑠+1)2 } − 𝐿−1 {(𝑠+1)2 } −𝑡 −1 =𝑒 𝐿 { +4 𝑠 +4 1 −𝑡 −1 𝑠 2 +22 }−𝑒 𝐿 { 𝑠 2 +22 } by shifting property = 𝑒 −2𝑡 cos 2 𝑡 − 𝑒 −𝑡 2 sin 2 𝑡(8) From (6),(7)(8) we get 10 10 𝐿−1 { 2 } = 𝐿−1 {(𝑠+2)(𝑠2 𝑠 +4𝑠+9𝑠+10 +2𝑠+5) = 2𝑒 −2𝑡 − 2 [𝑒 −𝑡 cos 2 𝑡 − = 2𝑒 =f(t) −2𝑡 − 2𝑒 −𝑡 cos 2 𝑡 + 𝑒 𝑒 −𝑡 2 −𝑡 Question #20 Find the Laplace transform 𝑡 1−𝑒 𝑢 𝐿{∫0 𝑢 𝑑𝑢} Answer: Step 1 Given: 𝑡 1−𝑒 𝑢 𝐿{∫0 𝑢 𝑑𝑢} Using Laplace Theorem 𝑡 𝑓(̅ 𝑠) = 𝐿{∫ 𝑓(𝑢)𝑑𝑢} = 𝑠 0 𝑡 1−𝑒 𝑢 = 𝐿{∫0 𝑢 1 1−𝑒 𝑢 𝑠 𝑢 𝑑𝑢} = 𝐿 [ Step 2 1 1 − 𝑒𝑢 𝐿[ ] 𝑠 𝑢 ∞ 1 ⇒ [∫𝑠 𝐹(𝑤)𝑑𝑤 ] ⇒ 𝑠 1 ∞ 1 1 𝑤 𝑤+1 ⇒ [∫𝑠 ( − 𝑠 1 ) 𝑑𝑤 ] ⇒ [(ln 𝑤 − ln(𝑤 + 1))∞ 𝑠 ] 𝑠 1 ⇒ [ln ( 𝑠 𝑤 ∞ ) ] 𝑤+1 𝑠 } 1 } = 2𝑒 −2𝑡 (7) 𝑠+2 𝑠+1−1 𝑠+1−1 𝐿−1 { 2 } = 𝐿−1 {(𝑠+1)2 } 𝑠 +2𝑠+1+4 +4 } = 2𝐿−1 { 𝑠 2 +2𝑠+5 𝑠+1−1 −1 𝑠+2 ] } sin 2 𝑡] sin 2 𝑡 1 ⇒ [− ln ( 𝑠 1 ⇒ [ln ( 𝑠 𝑠 𝑠+1 𝑠+1 𝑠 )] )]