Example 28 - Solve dx/dy + x/1 + y2 = tan-1 y / 1+y2 - Solving Linear differential equations - Equation given

Slide32.JPG
Slide33.JPG Slide34.JPG Slide35.JPG Slide36.JPG Slide37.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Example 28 (Introduction) Solve the differential equation 𝑑π‘₯/𝑑𝑦+π‘₯/(1+𝑦^2 )=tan^(βˆ’1)⁑𝑦/(1+𝑦^2 ) Now, 𝑑𝑦/𝑑π‘₯ = ((1 + 𝑦^2))/tan^(βˆ’1)β‘γ€–π‘¦βˆ’π‘₯γ€— Put F(x, y) = 𝑑𝑦/𝑑π‘₯ F(x, y) = (1 + 𝑦^2)/(tan^(βˆ’1)β‘π‘¦βˆ’π‘₯) F(πœ†x, πœ†y) = (1 + πœ†^2 𝑦^2)/(tan^(βˆ’1)β‘πœ†π‘¦βˆ’πœ†π‘₯) F(πœ†x, πœ†y) β‰  πœ†Β° F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method 𝑑𝑦/𝑑π‘₯ = ((1 + 𝑦^2))/(tan^(βˆ’1) y βˆ’ x) This is not of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 ∴ We need to find 𝑑π‘₯/𝑑𝑦 𝑑π‘₯/𝑑𝑦 = (tan^(βˆ’1)⁑𝑦 βˆ’ π‘₯)/(1 + 𝑦^2 ) 𝑑π‘₯/𝑑𝑦 = tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 ) βˆ’ π‘₯/(1 + 𝑦^2 ) 𝑑π‘₯/𝑑𝑦 + π‘₯/(1 + 𝑦^2 ) βˆ’ (tan^(βˆ’1)⁑𝑦 )/(1 + 𝑦^2 ) Differential equation is of the form 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 Thus, we solve question by integrating factor method taking 𝒅𝒙/π’…π’š Example 28 Solve the differential equation 𝑑π‘₯/𝑑𝑦+π‘₯/(1+𝑦^2 )=tan^(βˆ’1)⁑𝑦/(1+𝑦^2 ) IF = 𝑒^tan^(βˆ’1)⁑𝑦 Solution is x (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑𝑦+𝐢〗 x𝑒^tan^(βˆ’1)⁑𝑦 = ∫1β–’(tan^(βˆ’1)⁑〖𝑦 𝑒^tan^(βˆ’1)⁑𝑦 γ€— )/(1 + 𝑦^2 ) dy + C Let I = ∫1β–’(tan^(βˆ’1)⁑〖𝑦 𝑒^tan^(βˆ’1)⁑𝑦 γ€— )/(1 + 𝑦^2 ) dy Putting tan^(βˆ’1)⁑〖𝑦 γ€—= t 1/(1 + 𝑦^2 ) dy = dt Putting values of t & dt in I I = ∫1▒〖𝑑𝑒^𝑑 𝑑𝑑〗 I = t.∫1▒〖𝑒^𝑑 π‘‘π‘‘βˆ’βˆ«1β–’[1∫1▒〖𝑒^𝑑 𝑑𝑑〗] 𝑑𝑑〗 I = t𝑒^𝑑 βˆ’ ∫1▒〖𝑒^𝑑 𝑑𝑑〗 I = t𝑒^𝑑 βˆ’ 𝑒^𝑑 I = 𝑒^𝑑 (t βˆ’ 1) Putting value of t = tan^(βˆ’1)⁑𝑦 I = 𝑒^(tan^(βˆ’1)⁑𝑦 ) (tan^(βˆ’1) π‘¦βˆ’1) Putting value of I in (1) γ€–π‘₯𝑒〗^(tan^(βˆ’1)⁑𝑦 ) = 𝐼 + 𝐢 γ€–π‘₯𝑒〗^(tan^(βˆ’1) 𝑦 )= 𝑒^(tan^(βˆ’1) 𝑦 ) (tan^(βˆ’1)β‘π‘¦βˆ’1) + c Divide by tan^(βˆ’1) 𝑦 〖𝒙𝒆〗^(〖𝒕𝒂𝒏〗^(βˆ’πŸ) π’š )= (〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’šβˆ’πŸ) + c𝒆^(βˆ’γ€–π’•π’‚π’γ€—^(βˆ’πŸ) π’š ) Which 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 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail