Example 16 - Find general solution: ydx - (x + 2y2)dy = 0 - Examples

part 2 - Example 16 - Examples - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Example 16 - Examples - Serial order wise - Chapter 9 Class 12 Differential Equations

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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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