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Transcript

Example 16 Find the general solution of the differential equation 𝑦 𝑑π‘₯βˆ’(π‘₯+2𝑦^2 )𝑑𝑦=0 Given equation 𝑦 𝑑π‘₯βˆ’(π‘₯+2𝑦^2 )𝑑𝑦=0 𝑦 𝑑π‘₯=(π‘₯+2𝑦^2 )𝑑𝑦 π’…π’š/𝒅𝒙 = π’š/(𝒙 + πŸπ’š^𝟐 ) This is not of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 ∴ We find 𝒅𝒙/π’…π’š 𝑑π‘₯/𝑑𝑦 = (π‘₯ + 2𝑦^2)/𝑦 𝑑π‘₯/𝑑𝑦 = (π‘₯ )/𝑦 + (2𝑦^2)/𝑦 𝒅𝒙/π’…π’š βˆ’ (𝒙 )/π’š = 2y Differential equation is of the form 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 where P1 = (βˆ’1)/𝑦 & Q1 = 2y Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = 𝐞^∫1β–’γ€–(βˆ’πŸ)/π’š π’…π’š" " γ€— IF = e^(βˆ’log⁑𝑦 ) IF = e^log⁑(1/𝑦) IF = 𝟏/π’š Solution is x(IF) = ∫1β–’γ€–(𝑸×𝑰𝑭)π’…π’š+π‘ͺ γ€— x Γ— 1/𝑦=∫1β–’γ€–2𝑦×1/𝑦〗 𝑑𝑦+𝐢 𝒙/π’š = ∫1β–’γ€–πŸπ’…π’š+π‘ͺγ€— π‘₯/𝑦 = 2𝑦+𝐢 x = y (2y + C) x = 2y2 + Cy

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.