Slide13.JPG

Slide14.JPG
Slide15.JPG

Go Ad-free

Transcript

Example 6 Find the particular solution of the differential equation 𝑑𝑦/𝑑𝑥=−4𝑥𝑦^2 given that 𝑦=1 , 𝑤ℎ𝑒𝑛 𝑥=0Given differential equation , 𝑑𝑦/𝑑𝑥=−4𝑥𝑦^2 𝒅𝒚/𝒚^𝟐 = (−4 x) dx Integrating both sides. ∫1▒𝑑𝑦/𝑦^2 = ∫1▒〖−4𝑥 𝑑𝑥〗 ∫1▒𝒅𝒚/𝒚^𝟐 = −4 ∫1▒〖𝒙 𝒅𝒙〗 𝑦^(−2+1)/(−2+1) = −4.𝑥^2/2 + c − 𝟏/𝒚 = –2x2 + c y = (−1)/(−2𝑥2 + 𝑐) y = (−1)/(−(2𝑥2 − 𝑐)) y = 𝟏/(𝟐𝒙𝟐 − 𝒄) Given that at x = 0, y = 1 Putting x = 0, y = 1, in (1) 1 = 1/(2(0)^2 ) − c 1 = 1/(−𝐶) c = −1 Putting c = −1 in (1) y = 1/(2𝑥^2 ) −(−1) y = 𝟏/(𝟐𝒙^𝟐 + 𝟏) Hence, the particular solution of the equation is y = 𝟏/(𝟐𝒙^𝟐 + 𝟏)

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo