Ex 9.6, 8 - Chapter 9 Class 12 Differential Equations (Term 2)

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Ex 9.6, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : (1+𝑥^2 )𝑑𝑦+2𝑥𝑦 𝑑𝑥=cot⁡〖𝑥 𝑑𝑥(𝑥≠0)〗 Given equation (1 + x2)dy + 2xy dx = cot x dx Dividing both sides by dx (1 + x2)𝑑𝑦/𝑑𝑥 + 2xy 𝑑𝑥/𝑑𝑥 = cot x 𝑑𝑥/𝑑𝑥 (1 + x2)𝑑𝑦/𝑑𝑥 + 2xy = cot x Dividing both sides by (1 + x2) 𝑑𝑦/𝑑𝑥 + 2𝑥/((1 + 𝑥2)) y = cot⁡𝑥/((1 + 𝑥2)) Comparing (1) with 𝑑𝑦/𝑑𝑥 + Py = Q where P = 2𝑥/((1 + 𝑥^2)) & Q = cot⁡𝑥/((1 + 𝑥^2)) Finding Integrating factor, I.F I.F. = 𝑒^∫1▒〖𝑝 𝑑𝑥〗 I.F. = e^∫1▒〖2𝑥/((1 + 𝑥^2 ) ) 𝑑𝑥 〗 Let t = 1 + x2 dt = 2x dx …(1) I.F. = e^∫1▒〖𝑑𝑡/𝑡 〗 = elog t = t = (1 + x2) So, I.F = 1 + x2 Solution of the equation is y × I.F = ∫1▒〖𝑄×𝐼.𝐹.𝑑𝑥+𝐶〗 Putting values, y.(1 + x2) = ∫1▒cot⁡𝑥/((1 + 𝑥2)) × (1+𝑥2).dx + c y.(1 + x2) = ∫1▒〖cot⁡𝑥 𝑑𝑥〗+𝐶 y (1 + x2) = log |sin⁡𝑥 | + C Dividing by (1 + x2) y = (1 + x2)−1 log |𝐬𝐢𝐧⁡𝒙 |+𝑪(𝟏+"x2" )^(−𝟏) is the general solution of the given equation Note: This answer does not match with the answer of the book. If we have made any mistake, please comment