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Euler's and picard's
- The document discusses numerical methods for solving first order differential equations, namely Picard's method and Euler's method. - Picard's method involves iteratively replacing y with the previous approximation in the differential equation to obtain better approximations that converge to the solution. - Euler's method approximates the solution at the next point by the current value plus the rate of change times the change in x. This provides a first order approximation to the solution.
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Discusses the general first order differential equations and integral equations with initial conditions.
Introduces Picard's method with various approximations to solve differential equations through iterative substitutions.
Discusses the general first order differential equations and integral equations with initial conditions.Explains Euler's method for approximating solutions to differential equations, showcasing step-by-step calculations.
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Euler's and picard's
- 2. 𝐂𝐨𝐧𝐬𝐢𝐝𝐞𝐫 𝐭𝐡𝐞𝐠𝐞𝐧𝐞𝐫𝐚𝐥 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 𝐭𝐡𝐞𝟏 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: UsingPicard′ 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 𝐂𝐨𝐧𝐬𝐢𝐝𝐞𝐫𝐭𝐡𝐞 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: UsingEuler’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
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