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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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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