Solve the differential equation: ydx+(x-y 2 )dy=0

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Question 29 (Choice 1) Solve the differential equation: 𝑦𝑑π‘₯+(π‘₯βˆ’π‘¦^2 )𝑑𝑦=0 For equation 𝑦𝑑π‘₯+(π‘₯βˆ’π‘¦^2 )𝑑𝑦=0 We observe that we cannot use variable separation method Let’s try to put in the form π’…π’š/𝒅𝒙 + Py = Q or 𝒅𝒙/π’…π’š + P1 x = Q1 Now, y dx + (x βˆ’ y2) dy = 0 y dx = βˆ’ (x βˆ’ y2)dy π’…π’š/𝒅𝒙 = (βˆ’π’š)/(π’™βˆ’π’š^𝟐 ) This is not of the form 𝑑𝑦/𝑑π‘₯ + Py = Q Thus, let’s find 𝒅𝒙/π’…π’š 𝑑π‘₯/𝑑𝑦 = (𝑦^2 βˆ’ π‘₯)/𝑦 𝑑π‘₯/𝑑𝑦 = y βˆ’ π‘₯/𝑦 𝒅𝒙/π’…π’š + 𝒙/π’š = y Comparing with 𝒅𝒙/π’…π’š + P1 x = Q1 ∴ P1 = 1/𝑦 &. Q1 = y Finding Integrating factor, IF = 𝑒^∫1▒〖𝑝1 𝑑𝑦〗 = 𝑒^∫1▒𝑑𝑦/𝑦 = 𝒆^π’π’π’ˆβ‘π’š = y Solution is x (IF) = ∫1β–’γ€–(π‘ΈπŸ Γ— 𝑰𝑭)π’…π’š+𝒄〗 π‘₯𝑦=∫1▒〖𝑦 Γ— 𝑦 𝑑𝑦+𝑐〗 π’™π’š= ∫1β–’γ€–π’š^𝟐 π’…π’š+𝒄〗 π’™π’š= π’š^πŸ‘/πŸ‘+π‘ͺ

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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 and Computer Science at Teachoo