Ex 9.5, 11 - Find general solution: y dx + (x - y2) dy = 0 - Ex 9.5 part 2 - Ex 9.5, 11 - Ex 9.5 - Serial order wise - Chapter 9 Class 12 Differential Equations part 3 - Ex 9.5, 11 - Ex 9.5 - Serial order wise - Chapter 9 Class 12 Differential Equations

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Transcript

Ex 9.5, 11 For each of the differential equation find the general solution : 𝑦 𝑑π‘₯+(π‘₯βˆ’π‘¦^2 )𝑑𝑦=0 Step 1 : Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q or 𝑑π‘₯/𝑑𝑦 + P1 x = Q1, y dx + (x βˆ’ y2) dy = 0 y dx = βˆ’ (x βˆ’ y2)dy 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘¦)/(π‘₯βˆ’π‘¦^2 ) This is not of the form 𝑑𝑦/𝑑π‘₯ + Py = Q ∴ We 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 Step 2 : Find P1 and Q1 Comparing (1) with 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 Where P1 = 𝟏/π’š & Q1 = y Step 3 : Find Integrating factor, IF = 𝒆^∫1β–’γ€–π’‘πŸ π’…π’šγ€— = 𝑒^∫1▒𝑑𝑦/𝑦 = 𝑒^log⁑𝑦 = y Step 4 : Solution of the equation Solution is x (IF) = ∫1β–’γ€–(𝑄1×𝐼𝐹)𝑑𝑦+𝑐〗 xy = ∫1β–’γ€–π’šΓ—π’š π’…π’š+𝒄〗 xy = ∫1▒〖𝑦^2 𝑑𝑦+𝑐〗 xy = 𝑦^3/3+𝐢 x = 𝑦^3/3𝑦+𝐢/𝑦 x = π’š^𝟐/πŸ‘+π‘ͺ/π’š

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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