Example 28

Transcript

Example 28 (Introduction)Solve the differential equation𝑑𝑥/𝑑𝑦+𝑥/(1+𝑦^2 )=tan^(−1)⁡𝑦/(1+𝑦^2 ) Now, 𝑑𝑦/𝑑𝑥 = ((1 + 𝑦^2))/tan^(−1)⁡〖𝑦−𝑥〗 Put F(x, y) = 𝑑𝑦/𝑑𝑥 F(x, y) = (1 + 𝑦^2)/(tan^(−1)⁡𝑦−𝑥) F(𝜆x, 𝜆y) = (1 + 𝜆^2 𝑦^2)/(tan^(−1)⁡𝜆𝑦−𝜆𝑥) F(𝜆x, 𝜆y) ≠ 𝜆° F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method 𝑑𝑦/𝑑𝑥 = ((1 + 𝑦^2))/(tan^(−1) y − x) This is not of the form 𝑑𝑦/𝑑𝑥+𝑃𝑦=𝑄 ∴ We need to find 𝑑𝑥/𝑑𝑦 𝑑𝑥/𝑑𝑦 = (tan^(−1)⁡𝑦 − 𝑥)/(1 + 𝑦^2 ) 𝑑𝑥/𝑑𝑦 = tan^(−1)⁡𝑦/(1 + 𝑦^2 ) − 𝑥/(1 + 𝑦^2 ) 𝑑𝑥/𝑑𝑦 + 𝑥/(1 + 𝑦^2 ) − (tan^(−1)⁡𝑦 )/(1 + 𝑦^2 ) Differential equation is of the form 𝑑𝑥/𝑑𝑦 + P1 x = Q1 Thus, we solve question by integrating factor method taking 𝒅𝒙/𝒅𝒚 Example 28Solve the differential equation𝑑𝑥/𝑑𝑦+𝑥/(1+𝑦^2 )=tan^(−1)⁡𝑦/(1+𝑦^2 ) IF = 𝑒^tan^(−1)⁡𝑦 Solution is x (IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑦+𝐶〗 x𝑒^tan^(−1)⁡𝑦 = ∫1▒(tan^(−1)⁡〖𝑦 𝑒^tan^(−1)⁡𝑦 〗 )/(1 + 𝑦^2 ) dy + C Let I = ∫1▒(tan^(−1)⁡〖𝑦 𝑒^tan^(−1)⁡𝑦 〗 )/(1 + 𝑦^2 ) dy Putting tan^(−1)⁡〖𝑦 〗= t 1/(1 + 𝑦^2 ) dy = dt Putting values of t & dt in I I = ∫1▒〖𝑡𝑒^𝑡 𝑑𝑡〗 I = t.∫1▒〖𝑒^𝑡 𝑑𝑡−∫1▒[1∫1▒〖𝑒^𝑡 𝑑𝑡〗] 𝑑𝑡〗 I = t𝑒^𝑡 − ∫1▒〖𝑒^𝑡 𝑑𝑡〗 I = t𝑒^𝑡 − 𝑒^𝑡 I = 𝑒^𝑡 (t − 1) Putting value of t = tan^(−1)⁡𝑦 I = 𝑒^(tan^(−1)⁡𝑦 ) (tan^(−1) 𝑦−1) Putting value of I in (1) 〖𝑥𝑒〗^(tan^(−1)⁡𝑦 ) = 𝐼 + 𝐶 〖𝑥𝑒〗^(tan^(−1) 𝑦 )= 𝑒^(tan^(−1) 𝑦 ) (tan^(−1)⁡𝑦−1) + c Divide by tan^(−1) 𝑦 〖𝒙𝒆〗^(〖𝒕𝒂𝒏〗^(−𝟏) 𝒚 )= (〖𝒕𝒂𝒏〗^(−𝟏)⁡𝒚−𝟏) + c𝒆^(−〖𝒕𝒂𝒏〗^(−𝟏) 𝒚 ) Which is the required general solution