Euler's and picard's
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Numerical solution of ordinary differential equations, Euler's and Picard's methods
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Euler's and picard's
- 2. 𝐂𝐨𝐧𝐬𝐢𝐝𝐞𝐫 𝐭𝐡𝐞 𝐠𝐞𝐧𝐞𝐫𝐚𝐥 𝐟𝐢𝐫𝐬𝐭 𝐨𝐫𝐝𝐞𝐫 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝒅𝒚 𝒅𝒙 = 𝒇 𝒙, 𝒚 . − − − − − − 𝟏 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐢𝐭𝐢𝐚𝐥 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧 𝒚 𝒙 𝟎 = 𝒚 𝟎 𝐓𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟏 𝐜𝐚𝐧 𝐛𝐞 𝐰𝐫𝐢𝐭𝐞 𝐚𝐬 𝒅𝒚 = 𝒇 𝒙, 𝒚 𝒅𝒙 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐧𝐠 𝐨𝐧 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬 𝐰𝐢𝐭𝐡 𝐥𝐢𝐦𝐢𝐭𝐬 𝐟𝐨𝐫 𝒙 𝐚𝐧𝐝 𝒚, 𝐰𝐞 𝐠𝐞𝐭 𝒅𝒚 𝒚 𝒚 𝟎 = 𝒇 𝒙, 𝒚 𝒅𝒙 𝒙 𝒙 𝟎 (𝒚 − 𝒚 𝟎) = 𝒇 𝒙, 𝒚 𝒅𝒙 𝒙 𝒙 𝟎 𝒚 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝒅𝒙 𝒙 𝒙 𝟎 − − − − − (𝟐) 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 (2) is know as integral equation and can be solved by successive Approximation. MANIKANTA SATYALA
- 3. 𝐍𝐨𝐰 𝐭𝐡𝐞 𝟏 𝐬𝐭 𝐀𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐏𝐢𝐜𝐨𝐫𝐝′ 𝐬 𝐌𝐞𝐭𝐡𝐨𝐝 𝐁𝐲 𝐫𝐞𝐩𝐥𝐚𝐜𝐢𝐧𝐠 𝒚 𝐛𝐲 𝐲 𝟎 𝐢𝐧 𝒇 𝒙, 𝒚 𝐢𝐧 𝐑. 𝐇. 𝐒 𝐨𝐟 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟐 , 𝐰𝐞 𝐠𝐞𝐭 𝒚 𝟏 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝟎 𝒅𝒙 𝒙 𝒙 𝟎 𝐍𝐨𝐰 𝐭𝐡𝐞 𝟐 𝐧𝐝 𝐀𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐏𝐢𝐜𝐨𝐫𝐝′ 𝐬 𝐌𝐞𝐭𝐡𝐨𝐝 𝐁𝐲 𝐫𝐞𝐩𝐥𝐚𝐜𝐢𝐧𝐠 𝒚 𝐛𝐲 𝐲 𝟏 𝐢𝐧 𝒇 𝒙, 𝒚 𝐢𝐧 𝐑. 𝐇. 𝐒 𝐨𝐟 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟐 , 𝐰𝐞 𝐠𝐞𝐭 𝒚 𝟐 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝟏 𝒅𝒙 𝒙 𝒙 𝟎 𝐍𝐨𝐰 𝐭𝐡𝐞 𝐧𝐭𝐡 𝐀𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐏𝐢𝐜𝐨𝐫𝐝′ 𝐬 𝐌𝐞𝐭𝐡𝐨𝐝 𝐁𝐲 𝐫𝐞𝐩𝐥𝐚𝐜𝐢𝐧𝐠 𝒚 𝐛𝐲 𝐲 𝐧−𝟏 𝐢𝐧 𝒇 𝒙, 𝒚 𝐢𝐧 𝐑. 𝐇. 𝐒 𝐨𝐟 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟐 , 𝐰𝐞 𝐠𝐞𝐭 𝒚 𝒏 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝒏−𝟏 𝒅𝒙 𝒙 𝒙 𝟎 𝐒𝐭𝐨𝐩 𝐭𝐡𝐞 𝐩𝐫𝐨𝐜𝐞𝐬𝐬 𝐰𝐡𝐞𝐧 𝐲𝐨𝐮 𝐠𝐞𝐭 𝐭𝐰𝐨 𝐜𝐨𝐧𝐬𝐞𝐜𝐮𝐭𝐢𝐯𝐞 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟 𝒚 MANIKANTA SATYALA
- 4. MANIKANTA SATYALA Example: Using Picard′ s method solve 𝑑𝑦 𝑑𝑥 = 3 + 2𝑥𝑦 where 𝑦 0 = 1 for 𝑥 = 0.1 Solution: By Picard′ s Method 𝒚 𝒏 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝒏−𝟏 𝒅𝒙 𝒙 𝒙 𝟎 Here 𝑥0 = 0, 𝑦0 = 1, 𝑓 𝑥, 𝑦 = 3 + 2𝑥𝑦 1st Approximation: Put 𝑛 = 1 and 𝑦0 = 1 in 𝑓 𝑥, 𝑦 𝒚 𝟏 = 𝒚 𝟎 + 𝒇 𝒙, 𝒚 𝟎 𝒅𝒙 𝒙 𝒙 𝟎 ∴ 𝑦1 = 1 + 3𝑥 + 𝑥2 = 1.31(𝑎𝑡 𝑥 = 0.1) 2nd Approximation: Put 𝑛 = 2 and 𝑦1 = 1 + 3𝑥 + 𝑥2 in 𝑓 𝑥, 𝑦 𝒚 𝟐 = 𝟏 + 𝟑 + 𝟐𝒙 𝟏 + 𝟑𝒙 + 𝒙 𝟐 𝒅𝒙 𝒙 𝟎 ∴ 𝑦2 = 1 + 3𝑥 + 𝑥2 + 2𝑥3 + 𝑥4 2 = 1.31205(𝑎𝑡 𝑥 = 0.1) This is the approximation solution
- 5. MANIKANTA SATYALA 𝐂𝐨𝐧𝐬𝐢𝐝𝐞𝐫 𝐭𝐡𝐞 𝐠𝐞𝐧𝐞𝐫𝐚𝐥 𝐟𝐢𝐫𝐬𝐭 𝐨𝐫𝐝𝐞𝐫 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝒅𝒚 𝒅𝒙 = 𝒇 𝒙, 𝒚 , 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐢𝐭𝐢𝐚𝐥 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧 𝒚 𝒙 𝟎 = 𝒚 𝟎 𝐈𝐟 𝒚 𝒙 𝐢𝐬 𝐭𝐡𝐞 𝐞𝐱𝐚𝐜𝐭 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐨𝐟 𝒅𝒚 𝒅𝒙 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐓𝐚𝐲𝐥𝐨𝐫’𝐬 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟𝐨𝐫 𝒚(𝒙) 𝐢𝐬 𝒚 𝒙 = 𝒚(𝒙 𝟎) + 𝒙 − 𝒙 𝟎 𝟏! 𝒚′ 𝒙 𝟎 + 𝒙 − 𝒙 𝟎 𝟐 𝟐! 𝒚′′ 𝒙 𝟎 + .. .− − − 𝟏 𝐍𝐨𝐰 𝐬𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠 𝒉 = 𝒙 𝟏 – 𝒙 𝟎 𝐢𝐧 𝐞𝐪 𝟏 , 𝐰𝐞 𝐠𝐞𝐭 𝒚 𝒙 𝟏 = 𝒚 𝒙 𝟎 + 𝒉𝒚′ 𝒙 𝟎 + 𝒉 𝟐 𝟐! 𝒚′′(𝒙 𝟎) + ... 𝐈𝐟 𝒉 𝐢𝐬 𝐜𝐡𝐨𝐬𝐞𝐧 𝐬𝐦𝐚𝐥𝐥 𝐞𝐧𝐨𝐮𝐠𝐡 𝐭𝐡𝐞𝐧 𝐰𝐞 𝐦𝐚𝐲 𝐧𝐞𝐠𝐥𝐞𝐜𝐭 𝐭𝐡𝐞 𝐬𝐞𝐜𝐨𝐧𝐝 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡𝐞𝐫 𝐨𝐫𝐝𝐞𝐫 𝐭𝐞𝐫𝐦 𝐨𝐟 𝒉. 𝒚 𝟏 = 𝒚 𝟎 + 𝒉𝒇 𝒙 𝟎, 𝒚 𝟎 𝐖𝐡𝐢𝐜𝐡 𝐢𝐬 𝐄𝐮𝐥𝐞𝐫’𝐬 𝐟𝐢𝐫𝐬𝐭 𝐚𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐢𝐨𝐧. 𝐓𝐡𝐞 𝐠𝐞𝐧𝐞𝐫𝐚𝐥 𝐬𝐭𝐞𝐩 𝐟𝐨𝐫 𝐄𝐮𝐥𝐞𝐫 𝐦𝐞𝐭𝐡𝐨𝐝 𝐢𝐬 𝒚𝒊+𝟏 = 𝒚𝒊 + 𝒉𝒇(𝒙 𝒊 , 𝒚𝒊) 𝒘𝒉𝒆𝒓𝒆 𝒊 = 𝟎, 𝟏, 𝟐. ...
- 6. MANIKANTA SATYALA Problem: Using Euler’s method, find 𝑦 0.2 , given by 𝑑𝑦 𝑑𝑥 = 𝑦 − 2𝑥 𝑦 , 𝑦 0 = 1 and ℎ = 0.1 Solution: Here 𝑥0 = 0, 𝑦0 = 1 and 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 = 𝑦 − 2𝑥 𝑦 also ℎ = 0.1 1st Approximation: - 𝑦1 = 𝑦0 + ℎ 𝑓 𝑥0, 𝑦0 = 1 + 0.1 1 − 0 = 1.1 𝑥1 = 𝑥0 + ℎ = 0 + 0.1 = 0.1 2nd Approximation: - 𝑦2 = 𝑦1 + ℎ 𝑓 𝑥1,𝑦1 = 1.1 + 0.1 0.1 − 2 0.1 1.1 = 1.1918 𝑥2 = 𝑥1 + ℎ = 01 + 0.1 = 0.2 ∴ 𝑦 0.2 = 1.1918