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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 22 (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 ) π’…π’š/𝒅𝒙 = ((𝟏 + π’š^𝟐))/〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘γ€–π’š βˆ’ 𝒙〗 The variables cannot be separated. So variable separable method is not possible Now, 𝑑𝑦/𝑑π‘₯ = ((1 + 𝑦^2))/tan^(βˆ’1)β‘γ€–π‘¦βˆ’π‘₯γ€— Put F(x, y) = π’…π’š/𝒅𝒙 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 π’…π’š/𝒅𝒙+π‘·π’š=𝑸 ∴ We need to find 𝒅𝒙/π’…π’š 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 𝒅𝒙/π’…π’š + P1 x = Q1 Thus, we solve question by integrating factor method taking 𝒅𝒙/π’…π’š Example 22 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 𝒅𝒙/π’…π’š + 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’πŸ)β‘π’š Solution is x (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑𝑦+𝐢〗 x𝒆^〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’š = ∫1▒〖〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’š/(𝟏 + π’š^𝟐 )×𝒆^〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’š π’…π’šγ€— + C Let I = ∫1β–’γ€–tan^(βˆ’1)⁑𝑦/(1 + 𝑦^2 )×𝑒^tan^(βˆ’1)⁑𝑦 𝑑𝑦〗 Let 〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘γ€–π’š γ€—= t 1/(1 + 𝑦^2 ) dy = dt Putting values of t & dt in I I = ∫1▒〖𝒕𝒆^𝒕 𝒅𝒕〗 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" 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^𝒕 βˆ’ 𝒆^𝒕 I = 𝑒^𝑑 (t βˆ’ 1) Putting value of t = tan^(βˆ’1)⁑𝑦 I = 𝒆^(γ€–π­πšπ§γ€—^(βˆ’πŸ)β‘π’š ) (〖𝒕𝒂𝒏〗^(βˆ’πŸ) π’šβˆ’πŸ) 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 𝒆^(〖𝒕𝒂𝒏〗^(βˆ’πŸ) π’š ) 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.