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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.