Slide7.JPG

Slide8.JPG
Slide9.JPG

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


Transcript

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^(−log⁡𝑦 ) 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.