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Green function

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ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 1 GREEN FUNCTION Example 1:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 𝑥3 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 𝑥 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 𝑒3𝑧
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 2 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −𝑒 𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 + 𝑒2𝑡 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = − 𝑒 𝑡 2 (𝑒2𝑧|0 𝑡 ) + 𝑒2𝑡(𝑒 𝑧|0 𝑡 ) 𝑦(𝑡) = 𝑒3𝑡 2 + 𝑒 𝑡 2 − 𝑒2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦( 𝑥) = 𝑥3 2 + 𝑥 2 − 𝑥2
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 3 Example1-1:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 𝑥3 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 𝑥 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 𝐺(𝑥, 𝑧) = 𝐴(𝑧) 𝑥 + 𝐵(𝑧) 𝑥2 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑧 + 𝐵(𝑧) 𝑧2 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑧 𝐺,𝑥(𝑧, 𝑧) = 𝐴(𝑧) + 2𝐵(𝑧) 𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑧 + 2𝐵(𝑧) 𝑧 = 1 ∴ { 𝐴(𝑧) = −1 𝐵(𝑧) = 1 𝑧
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 4 𝐺(𝑡, 𝑧) = −𝑥 + 𝑥2 𝑧 , 𝑓(𝑧) = 𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑥 1 𝑦(𝑡) = −𝑥 ∫ 𝑧 𝑑𝑧 𝑥 1 + 𝑥2 ∫ 1 𝑑𝑧 𝑥 1 𝑦(𝑡) = − 𝑥 2 (𝑧2|0 𝑡 ) + 𝑥2(𝑧|0 𝑡 ) 𝑦( 𝑥) = 𝑥3 2 + 𝑥 2 − 𝑥2 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 𝑥^2′ , ′ 𝑦(1) = 0′ , ′ 𝐷𝑦(1) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^3 2⁄ + 𝑥 2⁄ − 𝑥^2 Example 2:- 𝑥2 𝑦,𝑥𝑥 + 𝑥𝑦,𝑥 − 𝑦 = 𝑥2 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = 𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 + 1 𝑥 𝑦,𝑥 − 1 𝑥2 𝑦 = 1 𝑃(𝑥) = 𝑒∫ 1 𝑥 𝑑𝑥 = 𝑥 𝑥𝑦,𝑥𝑥 + 𝑦,𝑥 − 1 𝑥 𝑦 = 1 𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 5 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) + 𝑅 − 1 = 0 𝑅2 − 1 = 0 ⇒ { 𝑅1 = 1 𝑅2 = −1 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒−𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒−𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒−𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−2𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 − 𝐵(𝑧) 𝑒−𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒−𝑧 − 𝐵(𝑧) 𝑒−𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−𝑧 2⁄ 𝐵(𝑧) = −𝑒 𝑧 2⁄ 𝐺(𝑡, 𝑧) = 1 2 𝑒−𝑧 𝑒 𝑡 − 1 2 𝑒 𝑧 𝑒−𝑡 , 𝑓(𝑧) = 𝑒2𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = 𝑒 𝑡 2 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 − 𝑒−𝑡 2 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = 𝑒 𝑡 2 (𝑒 𝑧|0 𝑡 ) − 𝑒−𝑡 6 (𝑒3𝑧|0 𝑡 ) 𝑦(𝑡) = 𝑒2𝑡 3 − 𝑒 𝑡 2 + 1 6𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝑥2 3 + 1 6𝑥 − 𝑥 2 𝑦( 𝑥) = 𝑥 ( 𝑥 3 + 1 6𝑥2 ) − 𝑥 2
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 6 Example 2-1:- 𝑥2 𝑦,𝑥𝑥 + 𝑥𝑦,𝑥 − 𝑦 = 𝑥2 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = 𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 + 1 𝑥 𝑦,𝑥 − 1 𝑥2 𝑦 = 1 𝑃(𝑥) = 𝑒∫ 1 𝑥 𝑑𝑥 = 𝑥 𝑥𝑦,𝑥𝑥 + 𝑦,𝑥 − 1 𝑥 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) + 𝑅 − 1 = 0 𝑅2 − 1 = 0 ⇒ { 𝑅1 = 1 𝑅2 = −1 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒−𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵 𝑥 𝐺(𝑥, 𝑧) = 𝐴(𝑧) 𝑥 + 𝐵(𝑧) 𝑥 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑧 + 𝐵(𝑧) 𝑧 = 0 ⇒ 𝐴(𝑧) = − 𝐵(𝑧) 𝑧2 𝐺,𝑥(𝑧, 𝑧) = 𝐴(𝑧) − 𝐵(𝑧) 𝑧2 = 1 ⇒ − 𝐵(𝑧) 𝑧2 − 𝐵(𝑧) 𝑧2 = 1
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 7 ∴ { 𝐴(𝑧) = 1 2 𝐵(𝑧) = − 𝑧2 2 𝐺(𝑡, 𝑧) = 𝑥 2 − 𝑧2 2𝑥 , 𝑓(𝑧) = 1 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑥 1 𝑦(𝑡) = 𝑥 2 ∫ 1 𝑑𝑧 𝑥 1 − 1 2𝑥 ∫ 𝑧2 𝑑𝑧 𝑥 1 𝑦(𝑡) = 𝑥 2 (𝑧|0 𝑡 ) − 1 6𝑥 (𝑧3|0 𝑡 ) 𝑦(𝑥) = 𝑥 2 (𝑥 − 1) − 1 6𝑥 (𝑥3 − 1) 𝑦(𝑥) = 𝑥2 3 + 1 6𝑥 − 𝑥 2 𝑦( 𝑥) = 𝑥 ( 𝑥 3 + 1 6𝑥2 ) − 𝑥 2 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 + 𝑥 ∗ 𝐷𝑦 − 𝑦 = 𝑥^2′ , ′ 𝑦(1) = 0′ , ′ 𝐷𝑦(1) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥 ∗ (𝑥 3⁄ + 1 (6 ∗ 𝑥^2)⁄ ) − 𝑥 2⁄
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 8 Example 3:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 2 𝑦(1) = 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 2
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 9 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −2𝑒 𝑡 ∫ 𝑒−𝑧 𝑑𝑧 𝑡 0 + 2𝑒2𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = 2𝑒 𝑡(𝑒−𝑧|0 𝑡 ) − 𝑒2𝑡(𝑒−2𝑧|0 𝑡 ) 2𝑒 𝑡(𝑒−𝑡 − 1) − 𝑒2𝑡(𝑒−2𝑡 − 1) 2 − 2𝑒 𝑡 − 1 + 𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒2𝑡 − 2𝑒 𝑡 + 1 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒2𝑡 − 2𝑒 𝑡 + 1 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥2 − 2𝑥 + 1 𝑦(1) = 𝐴 + 𝐵 + 1 − 2 + 1 = 0 ⇒ 𝐴 = −𝐵 𝑦(2) = 2𝐴 + 4𝐵 + 4 − 4 + 1 = 0 −2𝐵 + 4𝐵 = −1 ⇒ 𝐵 = − 1 2 , 𝐴 = 1 2 𝑦(𝑥) = 1 2 𝑥 − 1 2 𝑥2 + 𝑥2 − 2𝑥 + 1 𝑦( 𝑥) = 𝑥2 2 − 3𝑥 2 + 1 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2′ , ′ 𝑦(1) = 0′ , ′ 𝑦(2) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥2 2⁄ − 3 ∗ 𝑥 2⁄ + 1
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 10 Example 4:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 2𝑥3 𝑦(2) = 0 , 𝑦(3) = 6 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 2𝑒3𝑧
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 11 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −2𝑒 𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 + 2𝑒2𝑡 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −𝑒 𝑡(𝑒2𝑧|0 𝑡 ) + 2𝑒2𝑡(𝑒 𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −𝑒 𝑡(𝑒2𝑡 − 1) + 2𝑒2𝑡(𝑒 𝑡 − 1) 𝑦𝑝(𝑡) = −𝑒3𝑡 + 𝑒 𝑡 + 2𝑒3𝑡 − 2𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒3𝑡 − 2𝑒2𝑡 + 𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒3𝑡 − 2𝑒2𝑡 + 𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥3 − 2𝑥2 + 𝑥 𝑦(2) = 2𝐴 + 4𝐵 + 8 − 8 + 2 = 0 ⇒ 2𝐴 + 4𝐵 = −2 → (1) 𝑦(3) = 3𝐴 + 9𝐵 + 27 − 18 + 3 = 6 ⇒ 3𝐴 + 9𝐵 = −6 → (2) ∴ 𝐴 = 1 , 𝐵 = −1 𝑦(𝑥) = 𝑥 − 𝑥2 + 𝑥3 − 2𝑥2 + 𝑥 𝑦( 𝑥) = 𝑥3 − 3𝑥2 + 2𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2𝑥^3′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^3 − 3 ∗ 𝑥^2 + 2 ∗ 𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 12 Example 5:- 𝑥2 𝑦,𝑥𝑥 − 6 = 6𝑥 𝑦(1) = −1 , 𝑦(2) = 29 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 6 = 0 𝑅2 − 𝑅 − 6 = 0 ⇒ { 𝑅1 = 3 𝑅2 = −2 𝑦ℎ(𝑡) = 𝐴𝑒3𝑡 + 𝐵𝑒−2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒3𝑡 + 𝐵(𝑧) 𝑒−2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒3𝑧 + 𝐵(𝑧) 𝑒−2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−5𝑧 𝐺,𝑡(𝑧, 𝑧) = 3𝐴(𝑧) 𝑒3𝑧 − 2𝐵(𝑧) 𝑒−2𝑧 = 1 = −𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−3𝑧 5 𝐵(𝑧) = − 𝑒2𝑧 5 𝐺(𝑡, 𝑧) = 𝑒−3𝑧 5 𝑒3𝑡 − 𝑒2𝑧 5 𝑒−2𝑡 , 𝑓(𝑧) = 6𝑒 𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = 6 5 𝑒3𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 − 6 5 𝑒−2𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = − 6 10 𝑒3𝑡(𝑒−2𝑧|0 𝑡 ) − 6 15 𝑒−2𝑡(𝑒3𝑧|0 𝑡 )
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 13 𝑦𝑝(𝑡) = − 6 10 𝑒3𝑡(𝑒−2𝑡 − 1) − 6 15 𝑒−2𝑡(𝑒3𝑡 − 1) 𝑦𝑝(𝑡) = − 3 5 𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒 𝑡 + 2 5 𝑒−2𝑡 𝑦𝑝(𝑡) = −𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒−2𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒3𝑡 + 𝐵𝑒−2𝑡 − 𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒−2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥3 + 𝐵 𝑥2 − 𝑥 − 3 5 𝑥3 + 2 5𝑥2 𝑦(1) = 𝐴 + 𝐵 − 1 − 3 5 + 2 5 = −1 ⇒ 𝐴 = 1 5 − 𝐵 → (1) 𝑦(2) = 8𝐴 + 𝐵 4 − 2 − 24 5 + 2 20 = 29 = 8 ( 1 5 − 𝐵) + 𝐵 4 − 2 − 24 5 + 2 20 = 29 → × 20 ∴ 𝐵 = − 682 155 ⇒ 𝐴 = 713 155 𝑦(𝑥) = 713 155 𝑥3 − 682 155𝑥2 − 𝑥 − 3 5 𝑥3 + 2 5𝑥2 𝑦( 𝑥) = 4𝑥3 − 4 𝑥2 − 𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 6 ∗ 𝑦 = 6 ∗ 𝑥′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 4 ∗ 𝑥^3 − 4 𝑥^2⁄ − 𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 14 Example 6:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒2𝑧 + 𝐵(𝑧) 𝑒 𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−𝑧 𝐺,𝑡(𝑧, 𝑧) = 2𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 2𝐵(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒 𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−2𝑧 𝐵(𝑧) = −𝑒−𝑧 𝐺(𝑡, 𝑧) = 𝑒−2𝑧 𝑒2𝑡 − 𝑒−𝑧 𝑒 𝑡 , 𝑓(𝑧) = 6𝑒−𝑧
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 15 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = 6𝑒2𝑡 ∫ 𝑒−3𝑧 𝑑𝑧 𝑡 0 − 6𝑒 𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −2𝑒2𝑡(𝑒−3𝑧|0 𝑡 ) − 3𝑒 𝑡(𝑒−2𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −2𝑒−2𝑡(𝑒−3𝑡 − 1) + 𝑒 𝑡(𝑒2𝑡 − 1) 𝑦𝑝(𝑡) = −2𝑒−𝑡 + 2𝑒2𝑡 + 3𝑒−𝑡 + 3𝑒 𝑡 𝑦𝑝(𝑡) = 2𝑒2𝑡 + 𝑒−𝑡 + 3𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒2𝑡 + 𝐵𝑒 𝑡 + 2𝑒2𝑡 + 𝑒−𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 2𝑥2 + 1 𝑥 + 3𝑥 𝑦(1) = 𝐴 + 𝐵 + 2 + 1 + 3 = 1 ⇒ 𝐴 + 𝐵 = −5 → (1) 𝑦(2) = 4𝐴 + 2𝐵 + 8 + 1 2 + 6 = 0 ⇒ 2𝐴 + 4𝐵 = −14 → (2) ∴ 𝐴 = −2 , 𝐵 = −3 𝑦( 𝑥) = 1 𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 6 𝑥⁄ ′ , ′ 𝑦(1) = 1′ , ′ 𝑦(2) = 0.5′ ,′ 𝑥′ ) ≫ 𝑦 = 1 𝑥⁄
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 16 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 6𝑒−𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 6 (𝑠 + 1) 𝑦(𝑠) = 6 (𝑠 + 1)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 + 1)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 + 1) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 𝐵 + 𝐶𝑠2 − 𝐶𝑠 − 2𝑐 = 6 𝐴 = 1 , 𝐵 = 2 , 𝐶 = −3 𝑦(𝑠) = 1 (𝑠 + 1) + 2 (𝑠 − 2) − 3 (𝑠 − 1) 𝑦(𝑡) = 𝑒−𝑡 + 2𝑒2𝑡 − 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 2𝑥2 + 1 𝑥 − 3𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 17 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 2𝑥2 + 1 𝑥 − 3𝑥 𝑦(1) = 1 ⇒ 𝐴 + 𝐵 = 1 𝑦(2) = 1 2 ⇒ 4𝐴 + 2𝐵 = −2 𝐴 = −2 , 𝐵 = 3 𝑦( 𝑥) = 1 𝑥 Use variation of parameter method 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑤 = | 𝑒 𝑡 𝑒2𝑡 𝑒 𝑡 2𝑒2𝑡| = 𝑒3𝑡 𝑢1̀ = | 0 𝑒2𝑡 6𝑒−𝑡 2𝑒2𝑡| 𝑒3𝑡 = −6𝑒−2𝑡 𝑢1 = ∫ −6𝑒−2𝑡 𝑑𝑡 = 3𝑒−2𝑡
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 18 𝑢2̀ = | 𝑒 𝑡 0 𝑒 𝑡 6𝑒−𝑡| 𝑒3𝑡 = 6𝑒−3𝑡 𝑢2 = ∫ 6𝑒−3𝑡 𝑑𝑡 = −2𝑒−3𝑡 𝑦𝑝(𝑡) = 𝑢1 𝑦1 + 𝑢2 𝑦2 = 𝑒−𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒−𝑡 𝑙𝑒𝑡 𝑡 = ln(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 1 𝑥 𝑦(1) ⇒ 𝐴 = −𝐵 𝑦(2) ⇒ 𝐵 = 0 ∴ 𝐴 = 0 𝑦( 𝑥) = 1 𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 19 Example 7:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 12𝑥5 𝑦(1) = 0 , 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 12𝑒5𝑧
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 20 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −12𝑒 𝑡 ∫ 𝑒4𝑧 𝑑𝑧 𝑡 0 + 12𝑒2𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −3𝑒 𝑡(𝑒4𝑧|0 𝑡 ) + 4𝑒2𝑡(𝑒3𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −3𝑒 𝑡(𝑒4𝑡 − 1) + 4𝑒2𝑡(𝑒3𝑡 − 1) 𝑦𝑝(𝑡) = −3𝑒5𝑡 + 3𝑒 𝑡 + 4𝑒5𝑡 + 4𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒5𝑡 + 4𝑒2𝑡 + 3𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒5𝑡 + 4𝑒2𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥5 + 4𝑥2 + 3𝑥 𝑦(1) = 2𝐴 + 4𝐵 + 1 + 4 + 3 = 0 ⇒ 𝐴 + 𝐵 = −8 → (1) 𝑦(2) = 2𝐴 + 4𝐵 + 32 + 16 + 6 = 0 ⇒ 2𝐴 + 4𝐵 = −54 → (2) ∴ 𝐴 = 11 , 𝐵 = −19 𝑦(𝑥) = 11𝑥 − 19𝑥2 + 𝑥5 + 4𝑥2 + 3𝑥 𝑦( 𝑥) = 𝑥5 − 15𝑥2 + 14𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2 ∗ 𝑥^3′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^5 − 15 ∗ 𝑥^2 + 14 ∗ 𝑥
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 21 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 12𝑥5 𝑦(1) = 0 , 𝑦(2) = 0 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 𝑒5𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 12 (𝑠 − 5) 𝑦(𝑠) = 6 (𝑠 − 5)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 − 5)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 − 5) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 6𝐵𝑠 + 𝐵 + 𝐶𝑠2 − 7𝐶𝑠 + 10𝑐 = 12 𝐴 = 1 , 𝐵 = −4 , 𝐶 = 3 𝑦(𝑠) = 1 (𝑠 − 5) − 4 (𝑠 − 2) + 3 (𝑠 − 1) 𝑦(𝑡) = 𝑒5𝑡 − 4𝑒2𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 𝑥5 − 4𝑥2 + 3𝑥 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 𝑥5 − 4𝑥2 + 3𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 0
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 22 𝑦(2) = 0 ⇒ 4𝐴 + 2𝐵 = −22 𝐴 = −11 , 𝐵 = 11 𝑦( 𝑥) = 𝑥5 − 15𝑥2 + 14𝑥 Example 8:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6𝑥4 𝑦(1) = 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 23 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 6𝑒4𝑡 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −6𝑒 𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 + 6𝑒2𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = −2𝑒 𝑡(𝑒3𝑧|0 𝑡 ) + 3𝑒2𝑡(𝑒2𝑧|0 𝑡 ) 𝑦(𝑡) = −2𝑒 𝑡(𝑒3𝑡 − 1) + 3𝑒2𝑡(𝑒2𝑡 − 1) 𝑦(𝑡) = −2𝑒4𝑡 − 2𝑒 𝑡 + 3𝑒4𝑡 − 3𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒4𝑡 − 3𝑒2𝑡 − 2𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒4𝑡 − 3𝑒2𝑡 − 2𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥4 − 3𝑥2 − 2𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 4 𝑦(2) = 0 ⇒ 2𝐵 + 4𝐵 = 0 𝐴 = 8 , 𝐵 = −4 𝑦(𝑥) = 8𝑥 − 4𝑥2 + 𝑥4 − 3𝑥2 − 2𝑥 𝑦( 𝑥) = 𝑥4 − 7𝑥2 + 6𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2′ , ′ 𝑦(1) = 0′ , ′ 𝑦(2) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^2 2⁄ − 3 ∗ 𝑥 2⁄ + 1
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 24 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6𝑥4 𝑦(1) = 0 , 𝑦(2) = 0 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 6𝑒4𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 6 (𝑠 − 4) 𝑦(𝑠) = 6 (𝑠 − 4)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 − 4)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 − 4) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 5𝐵𝑠 + 4𝐵 + 𝐶𝑠2 − 6𝐶𝑠 + 8𝑐 = 6 𝐴 = 1 , 𝐵 = −3 , 𝐶 = 2 𝑦(𝑠) = 1 (𝑠 − 4) − 3 (𝑠 − 2) + 2 (𝑠 − 1) 𝑦(𝑡) = 𝑒4𝑡 − 3𝑒2𝑡 + 2𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 𝑥4 − 3𝑥2 + 2𝑥 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 𝑥4 − 3𝑥2 + 2𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 0
ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka 25 𝑦(2) = 0 ⇒ 4𝐴 + 2𝐵 = −8 𝐴 = −4 , 𝐵 = 4 𝑦( 𝑥) = 𝑥4 − 7𝑥2 + 6𝑥

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Green function

  • 1.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka1 GREEN FUNCTION Example 1:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 𝑥3 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 𝑥 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 𝑒3𝑧
  • 2.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka2 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −𝑒 𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 + 𝑒2𝑡 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = − 𝑒 𝑡 2 (𝑒2𝑧|0 𝑡 ) + 𝑒2𝑡(𝑒 𝑧|0 𝑡 ) 𝑦(𝑡) = 𝑒3𝑡 2 + 𝑒 𝑡 2 − 𝑒2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦( 𝑥) = 𝑥3 2 + 𝑥 2 − 𝑥2
  • 3.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka3 Example1-1:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 𝑥3 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 𝑥 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 𝐺(𝑥, 𝑧) = 𝐴(𝑧) 𝑥 + 𝐵(𝑧) 𝑥2 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑧 + 𝐵(𝑧) 𝑧2 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑧 𝐺,𝑥(𝑧, 𝑧) = 𝐴(𝑧) + 2𝐵(𝑧) 𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑧 + 2𝐵(𝑧) 𝑧 = 1 ∴ { 𝐴(𝑧) = −1 𝐵(𝑧) = 1 𝑧
  • 4.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka4 𝐺(𝑡, 𝑧) = −𝑥 + 𝑥2 𝑧 , 𝑓(𝑧) = 𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑥 1 𝑦(𝑡) = −𝑥 ∫ 𝑧 𝑑𝑧 𝑥 1 + 𝑥2 ∫ 1 𝑑𝑧 𝑥 1 𝑦(𝑡) = − 𝑥 2 (𝑧2|0 𝑡 ) + 𝑥2(𝑧|0 𝑡 ) 𝑦( 𝑥) = 𝑥3 2 + 𝑥 2 − 𝑥2 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 𝑥^2′ , ′ 𝑦(1) = 0′ , ′ 𝐷𝑦(1) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^3 2⁄ + 𝑥 2⁄ − 𝑥^2 Example 2:- 𝑥2 𝑦,𝑥𝑥 + 𝑥𝑦,𝑥 − 𝑦 = 𝑥2 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = 𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 + 1 𝑥 𝑦,𝑥 − 1 𝑥2 𝑦 = 1 𝑃(𝑥) = 𝑒∫ 1 𝑥 𝑑𝑥 = 𝑥 𝑥𝑦,𝑥𝑥 + 𝑦,𝑥 − 1 𝑥 𝑦 = 1 𝑥
  • 5.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka5 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) + 𝑅 − 1 = 0 𝑅2 − 1 = 0 ⇒ { 𝑅1 = 1 𝑅2 = −1 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒−𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒−𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒−𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−2𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 − 𝐵(𝑧) 𝑒−𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒−𝑧 − 𝐵(𝑧) 𝑒−𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−𝑧 2⁄ 𝐵(𝑧) = −𝑒 𝑧 2⁄ 𝐺(𝑡, 𝑧) = 1 2 𝑒−𝑧 𝑒 𝑡 − 1 2 𝑒 𝑧 𝑒−𝑡 , 𝑓(𝑧) = 𝑒2𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = 𝑒 𝑡 2 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 − 𝑒−𝑡 2 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = 𝑒 𝑡 2 (𝑒 𝑧|0 𝑡 ) − 𝑒−𝑡 6 (𝑒3𝑧|0 𝑡 ) 𝑦(𝑡) = 𝑒2𝑡 3 − 𝑒 𝑡 2 + 1 6𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝑥2 3 + 1 6𝑥 − 𝑥 2 𝑦( 𝑥) = 𝑥 ( 𝑥 3 + 1 6𝑥2 ) − 𝑥 2
  • 6.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka6 Example 2-1:- 𝑥2 𝑦,𝑥𝑥 + 𝑥𝑦,𝑥 − 𝑦 = 𝑥2 𝑦(1) = 𝑦,𝑥(1) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = 𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 + 1 𝑥 𝑦,𝑥 − 1 𝑥2 𝑦 = 1 𝑃(𝑥) = 𝑒∫ 1 𝑥 𝑑𝑥 = 𝑥 𝑥𝑦,𝑥𝑥 + 𝑦,𝑥 − 1 𝑥 𝑦 = 1 𝑥 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) + 𝑅 − 1 = 0 𝑅2 − 1 = 0 ⇒ { 𝑅1 = 1 𝑅2 = −1 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒−𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵 𝑥 𝐺(𝑥, 𝑧) = 𝐴(𝑧) 𝑥 + 𝐵(𝑧) 𝑥 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑧 + 𝐵(𝑧) 𝑧 = 0 ⇒ 𝐴(𝑧) = − 𝐵(𝑧) 𝑧2 𝐺,𝑥(𝑧, 𝑧) = 𝐴(𝑧) − 𝐵(𝑧) 𝑧2 = 1 ⇒ − 𝐵(𝑧) 𝑧2 − 𝐵(𝑧) 𝑧2 = 1
  • 7.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka7 ∴ { 𝐴(𝑧) = 1 2 𝐵(𝑧) = − 𝑧2 2 𝐺(𝑡, 𝑧) = 𝑥 2 − 𝑧2 2𝑥 , 𝑓(𝑧) = 1 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑥 1 𝑦(𝑡) = 𝑥 2 ∫ 1 𝑑𝑧 𝑥 1 − 1 2𝑥 ∫ 𝑧2 𝑑𝑧 𝑥 1 𝑦(𝑡) = 𝑥 2 (𝑧|0 𝑡 ) − 1 6𝑥 (𝑧3|0 𝑡 ) 𝑦(𝑥) = 𝑥 2 (𝑥 − 1) − 1 6𝑥 (𝑥3 − 1) 𝑦(𝑥) = 𝑥2 3 + 1 6𝑥 − 𝑥 2 𝑦( 𝑥) = 𝑥 ( 𝑥 3 + 1 6𝑥2 ) − 𝑥 2 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 + 𝑥 ∗ 𝐷𝑦 − 𝑦 = 𝑥^2′ , ′ 𝑦(1) = 0′ , ′ 𝐷𝑦(1) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥 ∗ (𝑥 3⁄ + 1 (6 ∗ 𝑥^2)⁄ ) − 𝑥 2⁄
  • 8.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka8 Example 3:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 2 𝑦(1) = 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 2
  • 9.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka9 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −2𝑒 𝑡 ∫ 𝑒−𝑧 𝑑𝑧 𝑡 0 + 2𝑒2𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = 2𝑒 𝑡(𝑒−𝑧|0 𝑡 ) − 𝑒2𝑡(𝑒−2𝑧|0 𝑡 ) 2𝑒 𝑡(𝑒−𝑡 − 1) − 𝑒2𝑡(𝑒−2𝑡 − 1) 2 − 2𝑒 𝑡 − 1 + 𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒2𝑡 − 2𝑒 𝑡 + 1 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒2𝑡 − 2𝑒 𝑡 + 1 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥2 − 2𝑥 + 1 𝑦(1) = 𝐴 + 𝐵 + 1 − 2 + 1 = 0 ⇒ 𝐴 = −𝐵 𝑦(2) = 2𝐴 + 4𝐵 + 4 − 4 + 1 = 0 −2𝐵 + 4𝐵 = −1 ⇒ 𝐵 = − 1 2 , 𝐴 = 1 2 𝑦(𝑥) = 1 2 𝑥 − 1 2 𝑥2 + 𝑥2 − 2𝑥 + 1 𝑦( 𝑥) = 𝑥2 2 − 3𝑥 2 + 1 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2′ , ′ 𝑦(1) = 0′ , ′ 𝑦(2) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥2 2⁄ − 3 ∗ 𝑥 2⁄ + 1
  • 10.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka10 Example 4:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 2𝑥3 𝑦(2) = 0 , 𝑦(3) = 6 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 2𝑒3𝑧
  • 11.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka11 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −2𝑒 𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 + 2𝑒2𝑡 ∫ 𝑒 𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −𝑒 𝑡(𝑒2𝑧|0 𝑡 ) + 2𝑒2𝑡(𝑒 𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −𝑒 𝑡(𝑒2𝑡 − 1) + 2𝑒2𝑡(𝑒 𝑡 − 1) 𝑦𝑝(𝑡) = −𝑒3𝑡 + 𝑒 𝑡 + 2𝑒3𝑡 − 2𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒3𝑡 − 2𝑒2𝑡 + 𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒3𝑡 − 2𝑒2𝑡 + 𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥3 − 2𝑥2 + 𝑥 𝑦(2) = 2𝐴 + 4𝐵 + 8 − 8 + 2 = 0 ⇒ 2𝐴 + 4𝐵 = −2 → (1) 𝑦(3) = 3𝐴 + 9𝐵 + 27 − 18 + 3 = 6 ⇒ 3𝐴 + 9𝐵 = −6 → (2) ∴ 𝐴 = 1 , 𝐵 = −1 𝑦(𝑥) = 𝑥 − 𝑥2 + 𝑥3 − 2𝑥2 + 𝑥 𝑦( 𝑥) = 𝑥3 − 3𝑥2 + 2𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2𝑥^3′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^3 − 3 ∗ 𝑥^2 + 2 ∗ 𝑥
  • 12.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka12 Example 5:- 𝑥2 𝑦,𝑥𝑥 − 6 = 6𝑥 𝑦(1) = −1 , 𝑦(2) = 29 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 6 = 0 𝑅2 − 𝑅 − 6 = 0 ⇒ { 𝑅1 = 3 𝑅2 = −2 𝑦ℎ(𝑡) = 𝐴𝑒3𝑡 + 𝐵𝑒−2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒3𝑡 + 𝐵(𝑧) 𝑒−2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒3𝑧 + 𝐵(𝑧) 𝑒−2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−5𝑧 𝐺,𝑡(𝑧, 𝑧) = 3𝐴(𝑧) 𝑒3𝑧 − 2𝐵(𝑧) 𝑒−2𝑧 = 1 = −𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−3𝑧 5 𝐵(𝑧) = − 𝑒2𝑧 5 𝐺(𝑡, 𝑧) = 𝑒−3𝑧 5 𝑒3𝑡 − 𝑒2𝑧 5 𝑒−2𝑡 , 𝑓(𝑧) = 6𝑒 𝑧 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = 6 5 𝑒3𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 − 6 5 𝑒−2𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = − 6 10 𝑒3𝑡(𝑒−2𝑧|0 𝑡 ) − 6 15 𝑒−2𝑡(𝑒3𝑧|0 𝑡 )
  • 13.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka13 𝑦𝑝(𝑡) = − 6 10 𝑒3𝑡(𝑒−2𝑡 − 1) − 6 15 𝑒−2𝑡(𝑒3𝑡 − 1) 𝑦𝑝(𝑡) = − 3 5 𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒 𝑡 + 2 5 𝑒−2𝑡 𝑦𝑝(𝑡) = −𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒−2𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒3𝑡 + 𝐵𝑒−2𝑡 − 𝑒 𝑡 − 3 5 𝑒3𝑡 + 2 5 𝑒−2𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥3 + 𝐵 𝑥2 − 𝑥 − 3 5 𝑥3 + 2 5𝑥2 𝑦(1) = 𝐴 + 𝐵 − 1 − 3 5 + 2 5 = −1 ⇒ 𝐴 = 1 5 − 𝐵 → (1) 𝑦(2) = 8𝐴 + 𝐵 4 − 2 − 24 5 + 2 20 = 29 = 8 ( 1 5 − 𝐵) + 𝐵 4 − 2 − 24 5 + 2 20 = 29 → × 20 ∴ 𝐵 = − 682 155 ⇒ 𝐴 = 713 155 𝑦(𝑥) = 713 155 𝑥3 − 682 155𝑥2 − 𝑥 − 3 5 𝑥3 + 2 5𝑥2 𝑦( 𝑥) = 4𝑥3 − 4 𝑥2 − 𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 6 ∗ 𝑦 = 6 ∗ 𝑥′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 4 ∗ 𝑥^3 − 4 𝑥^2⁄ − 𝑥
  • 14.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka14 Example 6:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒2𝑧 + 𝐵(𝑧) 𝑒 𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒−𝑧 𝐺,𝑡(𝑧, 𝑧) = 2𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 2𝐵(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒 𝑧 = 1 ∴ { 𝐴(𝑧) = 𝑒−2𝑧 𝐵(𝑧) = −𝑒−𝑧 𝐺(𝑡, 𝑧) = 𝑒−2𝑧 𝑒2𝑡 − 𝑒−𝑧 𝑒 𝑡 , 𝑓(𝑧) = 6𝑒−𝑧
  • 15.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka15 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = 6𝑒2𝑡 ∫ 𝑒−3𝑧 𝑑𝑧 𝑡 0 − 6𝑒 𝑡 ∫ 𝑒−2𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −2𝑒2𝑡(𝑒−3𝑧|0 𝑡 ) − 3𝑒 𝑡(𝑒−2𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −2𝑒−2𝑡(𝑒−3𝑡 − 1) + 𝑒 𝑡(𝑒2𝑡 − 1) 𝑦𝑝(𝑡) = −2𝑒−𝑡 + 2𝑒2𝑡 + 3𝑒−𝑡 + 3𝑒 𝑡 𝑦𝑝(𝑡) = 2𝑒2𝑡 + 𝑒−𝑡 + 3𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒2𝑡 + 𝐵𝑒 𝑡 + 2𝑒2𝑡 + 𝑒−𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 2𝑥2 + 1 𝑥 + 3𝑥 𝑦(1) = 𝐴 + 𝐵 + 2 + 1 + 3 = 1 ⇒ 𝐴 + 𝐵 = −5 → (1) 𝑦(2) = 4𝐴 + 2𝐵 + 8 + 1 2 + 6 = 0 ⇒ 2𝐴 + 4𝐵 = −14 → (2) ∴ 𝐴 = −2 , 𝐵 = −3 𝑦( 𝑥) = 1 𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 6 𝑥⁄ ′ , ′ 𝑦(1) = 1′ , ′ 𝑦(2) = 0.5′ ,′ 𝑥′ ) ≫ 𝑦 = 1 𝑥⁄
  • 16.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka16 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 6𝑒−𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 6 (𝑠 + 1) 𝑦(𝑠) = 6 (𝑠 + 1)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 + 1)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 + 1) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 𝐵 + 𝐶𝑠2 − 𝐶𝑠 − 2𝑐 = 6 𝐴 = 1 , 𝐵 = 2 , 𝐶 = −3 𝑦(𝑠) = 1 (𝑠 + 1) + 2 (𝑠 − 2) − 3 (𝑠 − 1) 𝑦(𝑡) = 𝑒−𝑡 + 2𝑒2𝑡 − 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 2𝑥2 + 1 𝑥 − 3𝑥
  • 17.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka17 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 2𝑥2 + 1 𝑥 − 3𝑥 𝑦(1) = 1 ⇒ 𝐴 + 𝐵 = 1 𝑦(2) = 1 2 ⇒ 4𝐴 + 2𝐵 = −2 𝐴 = −2 , 𝐵 = 3 𝑦( 𝑥) = 1 𝑥 Use variation of parameter method 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6 𝑥 𝑦(1) = 1 , 𝑦(2) = 1 2 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑤 = | 𝑒 𝑡 𝑒2𝑡 𝑒 𝑡 2𝑒2𝑡| = 𝑒3𝑡 𝑢1̀ = | 0 𝑒2𝑡 6𝑒−𝑡 2𝑒2𝑡| 𝑒3𝑡 = −6𝑒−2𝑡 𝑢1 = ∫ −6𝑒−2𝑡 𝑑𝑡 = 3𝑒−2𝑡
  • 18.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka18 𝑢2̀ = | 𝑒 𝑡 0 𝑒 𝑡 6𝑒−𝑡| 𝑒3𝑡 = 6𝑒−3𝑡 𝑢2 = ∫ 6𝑒−3𝑡 𝑑𝑡 = −2𝑒−3𝑡 𝑦𝑝(𝑡) = 𝑢1 𝑦1 + 𝑢2 𝑦2 = 𝑒−𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒−𝑡 𝑙𝑒𝑡 𝑡 = ln(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 1 𝑥 𝑦(1) ⇒ 𝐴 = −𝐵 𝑦(2) ⇒ 𝐵 = 0 ∴ 𝐴 = 0 𝑦( 𝑥) = 1 𝑥
  • 19.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka19 Example 7:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 12𝑥5 𝑦(1) = 0 , 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 12𝑒5𝑧
  • 20.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka20 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −12𝑒 𝑡 ∫ 𝑒4𝑧 𝑑𝑧 𝑡 0 + 12𝑒2𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 𝑦𝑝(𝑡) = −3𝑒 𝑡(𝑒4𝑧|0 𝑡 ) + 4𝑒2𝑡(𝑒3𝑧|0 𝑡 ) 𝑦𝑝(𝑡) = −3𝑒 𝑡(𝑒4𝑡 − 1) + 4𝑒2𝑡(𝑒3𝑡 − 1) 𝑦𝑝(𝑡) = −3𝑒5𝑡 + 3𝑒 𝑡 + 4𝑒5𝑡 + 4𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒5𝑡 + 4𝑒2𝑡 + 3𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒5𝑡 + 4𝑒2𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥5 + 4𝑥2 + 3𝑥 𝑦(1) = 2𝐴 + 4𝐵 + 1 + 4 + 3 = 0 ⇒ 𝐴 + 𝐵 = −8 → (1) 𝑦(2) = 2𝐴 + 4𝐵 + 32 + 16 + 6 = 0 ⇒ 2𝐴 + 4𝐵 = −54 → (2) ∴ 𝐴 = 11 , 𝐵 = −19 𝑦(𝑥) = 11𝑥 − 19𝑥2 + 𝑥5 + 4𝑥2 + 3𝑥 𝑦( 𝑥) = 𝑥5 − 15𝑥2 + 14𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2 ∗ 𝑥^3′ , ′ 𝑦(2) = 0′ , ′ 𝑦(3) = 6′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^5 − 15 ∗ 𝑥^2 + 14 ∗ 𝑥
  • 21.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka21 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 12𝑥5 𝑦(1) = 0 , 𝑦(2) = 0 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 𝑒5𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 12 (𝑠 − 5) 𝑦(𝑠) = 6 (𝑠 − 5)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 − 5)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 − 5) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 6𝐵𝑠 + 𝐵 + 𝐶𝑠2 − 7𝐶𝑠 + 10𝑐 = 12 𝐴 = 1 , 𝐵 = −4 , 𝐶 = 3 𝑦(𝑠) = 1 (𝑠 − 5) − 4 (𝑠 − 2) + 3 (𝑠 − 1) 𝑦(𝑡) = 𝑒5𝑡 − 4𝑒2𝑡 + 3𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 𝑥5 − 4𝑥2 + 3𝑥 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 𝑥5 − 4𝑥2 + 3𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 0
  • 22.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka22 𝑦(2) = 0 ⇒ 4𝐴 + 2𝐵 = −22 𝐴 = −11 , 𝐵 = 11 𝑦( 𝑥) = 𝑥5 − 15𝑥2 + 14𝑥 Example 8:- 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6𝑥4 𝑦(1) = 𝑦(2) = 0 Solution (𝑎 𝑜(𝑥)) = 𝑥2 , (𝑎 𝑜(𝑥)),𝑥 = 2𝑥 , (𝑎1(𝑥)) = −2𝑥 (𝑎 𝑜(𝑥)),𝑥 ≠ (𝑎1(𝑥)) ∴ 𝑛𝑜𝑡 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑦,𝑥𝑥 − 2 𝑥 𝑦,𝑥 + 2 𝑥2 𝑦 = 2 𝑥2 𝑃(𝑥) = 𝑒∫ − 2 𝑥 𝑑𝑥 = 1 𝑥2 1 𝑥2 𝑦,𝑥𝑥 − 2 𝑥3 𝑦,𝑥 + 2 𝑥4 𝑦 = 1 𝑥4 (𝑎 𝑜(𝑥)),𝑥 = (𝑎1(𝑥)) ∴ 𝑠𝑒𝑙𝑓 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 1 𝑅2 = 2 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝐺(𝑡, 𝑧) = 𝐴(𝑧) 𝑒 𝑡 + 𝐵(𝑧) 𝑒2𝑡 𝐺(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 𝐵(𝑧) 𝑒2𝑧 = 0 ⇒ 𝐴(𝑧) = −𝐵(𝑧) 𝑒 𝑧 𝐺,𝑡(𝑧, 𝑧) = 𝐴(𝑧) 𝑒 𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1 ⇒ − 𝐵(𝑧) 𝑒2𝑧 + 2𝐵(𝑧) 𝑒2𝑧 = 1
  • 23.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka23 ∴ { 𝐴(𝑧) = −𝑒−𝑧 𝐵(𝑧) = 𝑒−2𝑧 𝐺(𝑡, 𝑧) = −𝑒−𝑧 𝑒 𝑡 + 𝑒−2𝑧 𝑒2𝑡 , 𝑓(𝑧) = 6𝑒4𝑡 𝑦(𝑡) = ∫ 𝐺(𝑡, 𝑧) 𝑓(𝑧) 𝑑𝑧 𝑡 0 𝑦(𝑡) = −6𝑒 𝑡 ∫ 𝑒3𝑧 𝑑𝑧 𝑡 0 + 6𝑒2𝑡 ∫ 𝑒2𝑧 𝑑𝑧 𝑡 0 𝑦(𝑡) = −2𝑒 𝑡(𝑒3𝑧|0 𝑡 ) + 3𝑒2𝑡(𝑒2𝑧|0 𝑡 ) 𝑦(𝑡) = −2𝑒 𝑡(𝑒3𝑡 − 1) + 3𝑒2𝑡(𝑒2𝑡 − 1) 𝑦(𝑡) = −2𝑒4𝑡 − 2𝑒 𝑡 + 3𝑒4𝑡 − 3𝑒2𝑡 𝑦𝑝(𝑡) = 𝑒4𝑡 − 3𝑒2𝑡 − 2𝑒 𝑡 𝑦(𝑡) = 𝑦ℎ(𝑡) + 𝑦𝑝(𝑡) 𝑦(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 + 𝑒4𝑡 − 3𝑒2𝑡 − 2𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦(𝑥) = 𝐴𝑥 + 𝐵𝑥2 + 𝑥4 − 3𝑥2 − 2𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 4 𝑦(2) = 0 ⇒ 2𝐵 + 4𝐵 = 0 𝐴 = 8 , 𝐵 = −4 𝑦(𝑥) = 8𝑥 − 4𝑥2 + 𝑥4 − 3𝑥2 − 2𝑥 𝑦( 𝑥) = 𝑥4 − 7𝑥2 + 6𝑥 𝑀𝐴𝑇𝐿𝐴𝐵 | ≫ 𝑦 = 𝑑𝑠𝑜𝑙𝑣𝑒( ′ 𝑥^2 ∗ 𝐷2𝑦 − 2 ∗ 𝑥 ∗ 𝐷𝑦 + 2 ∗ 𝑦 = 2′ , ′ 𝑦(1) = 0′ , ′ 𝑦(2) = 0′ ,′ 𝑥′) ≫ 𝑦 = 𝑥^2 2⁄ − 3 ∗ 𝑥 2⁄ + 1
  • 24.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka24 Use Laplace transform 𝑥2 𝑦,𝑥𝑥 − 2𝑥𝑦,𝑥 + 2𝑦 = 6𝑥4 𝑦(1) = 0 , 𝑦(2) = 0 Solution 𝑙𝑒𝑡: 𝑥 = 𝑒 𝑡 𝑅(𝑅 − 1) − 2𝑅 + 2 = 0 𝑅2 − 3𝑅 + 2 = 0 ⇒ { 𝑅1 = 2 𝑅2 = 1 𝑦ℎ(𝑡) = 𝐴𝑒 𝑡 + 𝐵𝑒2𝑡 𝑦,𝑡𝑡 − 3𝑦,𝑡 + 2𝑦 = 6𝑒4𝑡 (𝑠2 − 3𝑠 + 2)𝑦(𝑠) = 6 (𝑠 − 4) 𝑦(𝑠) = 6 (𝑠 − 4)(𝑠 − 2)(𝑠 − 1) 6 (𝑠 − 4)(𝑠 − 2)(𝑠 − 1) = 𝐴 (𝑠 − 4) + 𝐵 (𝑠 − 2) + 𝐶 (𝑠 − 1) 𝐴𝑠2 − 3𝐴𝑠 + 2𝐴 + 𝐵𝑠2 − 5𝐵𝑠 + 4𝐵 + 𝐶𝑠2 − 6𝐶𝑠 + 8𝑐 = 6 𝐴 = 1 , 𝐵 = −3 , 𝐶 = 2 𝑦(𝑠) = 1 (𝑠 − 4) − 3 (𝑠 − 2) + 2 (𝑠 − 1) 𝑦(𝑡) = 𝑒4𝑡 − 3𝑒2𝑡 + 2𝑒 𝑡 𝑙𝑒𝑡: 𝑡 = 𝑙𝑛(𝑥) 𝑦𝑝(𝑥) = 𝑥4 − 3𝑥2 + 2𝑥 𝑦(𝑥) = 𝑦ℎ(𝑥) + 𝑦𝑝(𝑥) 𝑦(𝑥) = 𝐴𝑥2 + 𝐵𝑥 + 𝑥4 − 3𝑥2 + 2𝑥 𝑦(1) = 0 ⇒ 𝐴 + 𝐵 = 0
  • 25.
    ADVANCED MATHEMATICS By.Eng. Hamza-Mahmoud-Dahoka25 𝑦(2) = 0 ⇒ 4𝐴 + 2𝐵 = −8 𝐴 = −4 , 𝐵 = 4 𝑦( 𝑥) = 𝑥4 − 7𝑥2 + 6𝑥