Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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Last updated at Aug. 14, 2023 by Teachoo
Example 16 Find the general solution of the differential equation 𝑦 𝑑𝑥−(𝑥+2𝑦^2 )𝑑𝑦=0 Given equation 𝑦 𝑑𝑥−(𝑥+2𝑦^2 )𝑑𝑦=0 𝑦 𝑑𝑥=(𝑥+2𝑦^2 )𝑑𝑦 𝒅𝒚/𝒅𝒙 = 𝒚/(𝒙 + 𝟐𝒚^𝟐 ) This is not of the form 𝑑𝑦/𝑑𝑥+𝑃𝑦=𝑄 ∴ We find 𝒅𝒙/𝒅𝒚 𝑑𝑥/𝑑𝑦 = (𝑥 + 2𝑦^2)/𝑦 𝑑𝑥/𝑑𝑦 = (𝑥 )/𝑦 + (2𝑦^2)/𝑦 𝒅𝒙/𝒅𝒚 − (𝒙 )/𝒚 = 2y Differential equation is of the form 𝑑𝑥/𝑑𝑦 + P1 x = Q1 where P1 = (−1)/𝑦 & Q1 = 2y Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = 𝐞^∫1▒〖(−𝟏)/𝒚 𝒅𝒚" " 〗 IF = e^(−log𝑦 ) IF = e^log(1/𝑦) IF = 𝟏/𝒚 Solution is x(IF) = ∫1▒〖(𝑸×𝑰𝑭)𝒅𝒚+𝑪 〗 x × 1/𝑦=∫1▒〖2𝑦×1/𝑦〗 𝑑𝑦+𝐶 𝒙/𝒚 = ∫1▒〖𝟐𝒅𝒚+𝑪〗 𝑥/𝑦 = 2𝑦+𝐶 x = y (2y + C) x = 2y2 + Cy