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

Transcript

Example 28 (Introduction) Solve the differential equation 𝑑π‘₯/𝑑𝑦+π‘₯/(1+𝑦^2 )=tan^(βˆ’1)⁑𝑦/(1+𝑦^2 ) 𝑑π‘₯/𝑑𝑦+π‘₯/(1 + 𝑦^2 )=tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 ) 𝑑π‘₯/𝑑𝑦=tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 ) – π‘₯/(1 + 𝑦^2 ) 𝑑π‘₯/𝑑𝑦=(tan^(βˆ’1)⁑𝑦 βˆ’ π‘₯)/(1 + 𝑦^2 ) 𝑑𝑦/𝑑π‘₯ = ((1 + 𝑦^2))/tan^(βˆ’1)⁑〖𝑦 βˆ’ π‘₯γ€— The variables cannot be separated. So variable separable method is not possible Now, 𝑑𝑦/𝑑π‘₯ = ((1 + 𝑦^2))/tan^(βˆ’1)⁑〖𝑦 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‘₯)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 ) 𝑑π‘₯/𝑑𝑦+π‘₯/(1 + 𝑦^2 )=tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 ) Differential equation is of the form 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 where P1 = 1/(1 + 𝑦^2 ) & Q1 = (tan^(βˆ’1)⁑𝑦 )/(1 + 𝑦^2 ) Now, IF = 𝑒^∫1▒〖𝑃_1 𝑑𝑦〗 IF = 𝑒^∫1β–’γ€–1/(1 + 𝑦^2 ) 𝑑𝑦〗 IF = 𝑒^tan^(βˆ’1)⁑𝑦 Solution is x (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑𝑦+𝐢〗 x𝑒^tan^(βˆ’1)⁑𝑦 = ∫1β–’γ€–tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 )×𝑒^tan^(βˆ’1)⁑𝑦 𝑑𝑦〗 + C Let I = ∫1β–’γ€–tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 )×𝑒^tan^(βˆ’1)⁑𝑦 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 Integrating by parts with ∫1▒〖𝑓(𝑑) 𝑔(𝑑) 𝑑𝑑=𝑓(𝑑) ∫1▒〖𝑔(𝑑) 𝑑𝑑 βˆ’βˆ«1β–’γ€–[𝑓^β€² (𝑑) ∫1▒〖𝑔(𝑑) 𝑑𝑑] 𝑑𝑑〗〗〗〗 Take f (t) = t & g (t) = 𝑒^"t" Divide by 𝑒^(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑦 ) γ€–π‘₯𝑒〗^(tan^(βˆ’1) 𝑦 )/𝑒^tan^(βˆ’1)⁑〖𝑦 γ€— = (𝑒^(tan^(βˆ’1) 𝑦) (tan^(βˆ’1)⁑𝑦 βˆ’ 1))/𝑒^tan^(βˆ’1)⁑〖𝑦 γ€— + 𝑐/𝑒^tan^(βˆ’1)⁑〖𝑦 γ€— 𝒙 = (〖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

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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.