Slide2.JPG

Slide3.JPG
Slide4.JPG Slide5.JPG

  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Misc 7 Show that the general solution of the differential equation 𝑑𝑦/𝑑π‘₯+(𝑦^2+𝑦+1)/(π‘₯^2+π‘₯+1)=0 is given by (π‘₯+𝑦+1)=A(1βˆ’π‘₯βˆ’π‘¦βˆ’2π‘₯𝑦), where A is parameter. 𝑑𝑦/𝑑π‘₯+(𝑦^2 + 𝑦 + 1)/(π‘₯^2 + π‘₯ + 1) = 0 𝑑𝑦/𝑑π‘₯=(βˆ’(𝑦^2 + 𝑦 + 1))/(π‘₯^2 + π‘₯ + 1) 𝑑𝑦/(𝑦^2 + 𝑦 + 1)=(βˆ’π‘‘π‘₯)/(π‘₯^2 + π‘₯ + 1) 𝑑𝑦/(𝑦^2 +2(1/2)𝑦 + (1/2)^2βˆ’ (1/2)^2+ 1)=(βˆ’π‘‘π‘₯)/(π‘₯^2 + 2(1/2)π‘₯ + (1/2)^2βˆ’ (1/2)^2+ 1) 𝑑𝑦/((𝑦 + 1/2)^2+ 3/4)=(βˆ’π‘‘π‘₯)/((π‘₯ + 1/2)^2+ 3/4) 𝑑𝑦/((𝑦 + 1/2)^2+ (√3/2)^2 )=(βˆ’π‘‘π‘₯)/((π‘₯ + 1/2)^2+ (√3/2)^2 ) Integrating both sides ∫1▒𝑑𝑦/((𝑦 + 1/2)^2 +(√3/2)^2 ) = βˆ’ ∫1▒𝑑π‘₯/((π‘₯ + 1/2)^2 +(√3/2)^2 ) 2/√3 tanβˆ’1 ((𝑦 + 1/2)/(√3/2)) = (βˆ’2)/√3 tanβˆ’1 ((π‘₯ + 1/2)/(√3/2)) + C 2/√3 ["tanβˆ’1 " ((2𝑦 + 1)/√3)" + tanβˆ’1 " ((2π‘₯ + 1)/√3)] = C (∫1▒𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=1/π‘Ž "tanβˆ’1" π‘₯/π‘Ž) (Using tanβˆ’1 A + tanβˆ’1 B = tanβˆ’1 ((𝐴 + 𝐡)/(1 βˆ’ 𝐴𝐡)) ) 2/√3 "tanβˆ’1" ⌈((2𝑦 + 1)/√3 + (2π‘₯ + 1)/√3)/(1 βˆ’ (2𝑦 βˆ’ 1)/√3 Γ—(2π‘₯ + 1)/√3 )βŒ‰=𝐢 "tanβˆ’1" [((2𝑦 + 1 + 2π‘₯ + 1)/√3)/(1 βˆ’ ((2𝑦 + 1)(2π‘₯ + 1))/√3)] = √3/2 𝐢 ((2𝑦 + 1 + 2π‘₯ + 1)/√3)/(1 βˆ’ ((2𝑦 + 1)(2π‘₯ + 1))/3) = tan (√3/2 𝐢) ((2𝑦 + 2π‘₯ +2)/√3)/((3 βˆ’ (2𝑦 + 1)(2π‘₯ + 1))/3) = C1 (Where C1 = tan (√3/2 𝐢)) (√3(2𝑦 + 2π‘₯ + 2))/(3 βˆ’ (4π‘₯𝑦 + 2𝑦 + 2π‘₯ + 1) ) = C 2√3 (x + y + 1) = C1 |3βˆ’4π‘₯π‘¦βˆ’2π‘₯βˆ’2π‘¦βˆ’1| 2√3 (x + y + 1) = C1 |2βˆ’4π‘₯π‘¦βˆ’2π‘₯βˆ’2𝑦| 2√3 (x + y + 1) = C1 Γ— 2 |1βˆ’π‘₯βˆ’π‘¦βˆ’π‘₯𝑦| βˆšπŸ‘ (x + y + 1) = C1 ( 1 βˆ’ x βˆ’ y βˆ’ 2xy) is the required general solution

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.