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Lecture5_Laplace_ODE.pdf

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Lecture5_Laplace_ODE.pdf

  1. 1. Ch3: Inverse Laplace Transforms + Solution of Linear Differential Equations Lecture 5
  2. 2. Properties of Inverse Laplace Transforms ‫العكسية‬ ‫البالس‬ ‫تحويالت‬ ‫خواص‬ 3 ℒ−1 𝑑 𝑑𝑠 𝐹 𝑠 = −𝑡𝑓 𝑡 4 ℒ−1 න 𝑠 ∞ 𝐹 𝑠 𝑑𝑠 = 𝑓 𝑡 𝑡 5 ℒ−1 𝐹 𝑠 𝑒−𝑎𝑠 = 𝑓 𝑡 − 𝑎 𝑢(𝑡 − 𝑎) ℒ−1 𝑐1𝐹1(s) ± 𝑐2𝐹2(s) = 𝑐1𝑓1 𝑡 ± 𝑐2𝑓2 𝑡 1 ℒ−1 𝐹 𝑠 ∓ 𝑎 = 𝑒±𝑎𝑡𝑓 𝑡 2 ‫المحاضرة‬ ‫السابقة‬
  3. 3. Find inverse Laplace of 𝐹 𝑠 = 3 𝑠 − 1 2 + 9 𝑒−4𝑠 Solution: ℒ−1 3 𝑠 − 1 2 + 9 = sin 3𝑡 𝑒𝑡 ∴ ℒ−1 3 𝑠 − 1 2 + 9 𝑒−4𝑠 = sin 3 𝑡 − 4 𝑒(𝑡−4)𝑢(𝑡 − 4) 𝐹 𝑠 = 1 𝑠 − 3 𝑒−2𝑠 Solution: ℒ−1 1 𝑠 − 3 = 𝑒3𝑡 ∴ ℒ−1 1 𝑠 − 3 𝑒−2𝑠 = 𝑒3(𝑡−2) 𝑢(𝑡 − 2)
  4. 4. Find inverse Laplace of 𝐹 𝑠 = 𝒔 𝒔𝟐 + 2𝒔 + 2 𝑒−3𝑠 Solution: By applying complete square (‫مربع‬ ‫)إكمال‬ ℒ−1 𝑠 𝑠 + 1 2 + 1 = ℒ−1 𝑠 + 1 − 1 𝑠 + 1 2 + 1 = ℒ−1 𝑠 + 1 𝑠 + 1 2 + 1 − ℒ−1 1 𝑠 + 1 2 + 1 = 𝑒−𝑡 cos 𝑡 − 𝑒−𝑡 sin 𝑡 = 𝑒−𝑡 cos 𝑡 − sin(𝑡) ℒ−1 𝑠 𝑠 + 1 2 + 1 𝑒−3𝑠 = 𝑒−(𝑡−3) cos 𝑡 − 3 − sin 𝑡 − 3 𝑢(𝑡 − 3)
  5. 5. Find inverse Laplace of 𝑭 𝒔 = 1 𝒔 + 1 𝒔𝟐 + 1 𝑒−𝜋𝑠 Solution: 1 𝒔 + 1 𝒔𝟐 + 1 = 𝑨 𝒔 + 𝟏 + 𝑩𝒔 + 𝑪 𝒔𝟐 + 𝟏 Multiplying both sides by 𝑠 + 1 (𝑠2 + 1) then Put 𝑠 = −1 → 𝐴 = 0.5, Equating the coefficient of 𝑠2 → 𝐵 = −0.5 Equating the coefficient of 𝑠0 → 𝐶 = 0.5 ℒ−1 1 𝒔 + 1 𝒔𝟐 + 1 = ℒ−1 0.5 𝑠 + 1 + −0.5𝑠 + 0.5 𝑠2 + 1 = 0.5𝑒−𝑡 − 0.5 cos 𝑡 + 0.5sin(𝑡) ℒ−1 1 𝒔 + 1 𝒔𝟐 + 1 𝑒−𝜋𝑠 = 0.5𝑒−(𝑡−𝜋) − 0.5 cos 𝑡 − 𝜋 + 0.5 sin 𝑡 − 𝜋 𝑢(𝑡 − 𝜋)
  6. 6. Solution of linear differential equations ℒ 𝑦 𝑡 = 𝑌 𝑠 ℒ 𝑦′ 𝑡 = 𝑠𝑌 𝑠 − 𝑦(0) ℒ 𝑦′′ 𝑡 = 𝑠2𝑌 𝑠 − 𝑠𝑦(0) − 𝑦′(0) ℒ 𝑦′′′ 𝑡 = 𝑠3𝑌 𝑠 − 𝑠2𝑦 0 − 𝑠𝑦′ 0 − 𝑦′′(0) and ℒ ‫׬‬ 0 𝑡 𝑦 𝑡 𝑑𝑡 = 𝑌(𝑠) 𝒔 Proof: ℒ 𝑦′ 𝑡 = න 0 ∞ 𝑦′ 𝑡 𝑒−𝑠𝑡𝑑𝑡 = 𝑦 𝑡 𝑒−𝑠𝑡 ቤ ∞ 0 + 𝑠 න 0 ∞ 𝑦 𝑡 𝑒−𝑠𝑡𝑑𝑡 ∴ ℒ 𝑦′ 𝑡 = −𝑦 0 + 𝑠𝑌 𝑠
  7. 7. 𝒚′ + 𝒚 = 𝒆−𝒕, 𝒚 𝟎 = 𝟏 Solution: 1 - ‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬ ℒ 𝒚′ + 𝒚 = ℒ 𝑒−𝑡 𝑠𝑌 𝑠 − ถ 𝑦(0) 1 + 𝑌 𝑠 = 1 𝑠 + 1 𝑠𝑌 𝑠 − 1 + 𝑌 𝑠 = 1 𝑠 + 1 2 - ‫نوجد‬ 𝒀(𝒔) ‫في‬ ‫طرف‬ 𝑠 + 1 𝑌 𝑠 = 1 𝑠 + 1 + 1 ⇒ 𝑌 𝑠 = 1 𝑠 + 1 2 + 1 𝑠 + 1 3 - ‫ناخذ‬ ‫البالس‬ ‫معكوس‬ ‫للطرفين‬ ℒ−1 𝑌 𝑠 = ℒ−1 1 𝑠 + 1 2 + 1 𝑠 + 1 ‫خواص‬+ ‫الجدول‬ ⇒ 𝑦 𝑡 = 𝑡𝑒−𝑡 + 𝑒−𝑡 = (𝑡 + 1)𝑒−𝑡 Solve using Laplace transforms
  8. 8. 𝒚′′ + 𝒚 = 𝟏, 𝒚 𝟎 = −𝟏, 𝒚′ 𝟎 = 𝟎 Solution: 1 - ‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬ ℒ 𝒚′′ + 𝒚 = ℒ 1 𝑠2 𝑌 𝑠 − 𝑠 ถ 𝑦(0) −1 − 𝑦′ (0) 0 + 𝑌 𝑠 = 1 𝑠 𝑠2𝑌 𝑠 + 𝑠 + 𝑌 𝑠 = 1 𝑠 2 - ‫نوجد‬ 𝒀(𝒔) ‫طرف‬ ‫في‬ (𝑠2 + 1)𝑌 𝑠 = 1 𝑠 − 𝑠 = 1 − 𝑠2 𝑠 ⇒ 𝑌 𝑠 = 1 − 𝑠2 𝑠(𝑠2 + 1) 3 - ‫ناخذ‬ ‫البالس‬ ‫معكوس‬ ‫للطرفين‬ ℒ−1 𝑌 𝑠 = ℒ−1 1 − 𝑠2 𝑠(𝑠2 + 1) 𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = ℒ−1 ฎ 𝐴 1 𝑠 + ฎ 𝐵 −2 𝑠 + ฎ 𝐶 0 (𝑠2 + 1) ⇒ 𝑦 𝑡 = 1 − 2𝑐𝑜𝑠 𝑡 Solve using Laplace transforms
  9. 9. 𝒚′ − න 0 𝑡 𝑦 𝑡 𝑑𝑡 = 𝒖(𝒕 − 𝟑), 𝒚 𝟎 = 𝟑, Solution: 1 - ‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬ ℒ 𝒚′ − න 0 𝑡 𝑦 𝑡 𝑑𝑡 = ℒ 𝑢(𝑡 − 3) 𝑠𝑌 𝑠 − ถ 𝑦 0 3 − 𝑌 𝑠 𝑠 = 1 𝑠 𝑒−3𝑠 𝑠 − 1 𝑠 𝑌 𝑠 = 1 𝑠 𝑒−3𝑠 + 3 2 - ‫نوجد‬ 𝒀(𝒔) ‫طرف‬ ‫في‬ 𝑠2 − 1 𝑠 𝑌 𝑠 = 1 𝑠 𝑒−3𝑠 + 3 ⇒ 𝑌 𝑠 = 1 𝑠2 − 1 𝑒−3𝑠 + 3𝑠 𝑠2 − 1 3 - ‫ناخذ‬ ‫البالس‬ ‫معكوس‬ ‫للطرفين‬ ℒ−1 𝑌 𝑠 = ℒ−1 1 𝑠2 − 1 𝑒−3𝑠 + 3𝑠 𝑠2 − 1 ⇒ 𝑦 𝑡 = sinh 𝑡 − 3 𝑢 𝑡 − 3 + 3cosh 𝑡 Solve using Laplace transforms

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