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Example 28 - Solve differential equation dx/dy + x/1+y^2 = tan-1 y /1

Example 28 - Chapter 9 Class 12 Differential Equations - Part 2
Example 28 - Chapter 9 Class 12 Differential Equations - Part 3
Example 28 - Chapter 9 Class 12 Differential Equations - Part 4
Example 28 - Chapter 9 Class 12 Differential Equations - Part 5
Example 28 - Chapter 9 Class 12 Differential Equations - Part 6
Example 28 - Chapter 9 Class 12 Differential Equations - Part 7

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.