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Ex 9.6, 8 - Find general solution: (1 + x2) dy + 2xy dx - Ex 9.6

Ex 9.6, 8 - Ex 9.6, 8For each of the differential equation given in Exercises 1 to 12, find the general solutio - Solving Linear differential equations - Equation given
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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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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+π‘₯^2 )𝑑𝑦+2π‘₯𝑦 𝑑π‘₯=cot⁑〖π‘₯ 𝑑π‘₯(π‘₯β‰ 0)γ€— Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q (1 + x2)dy + 2xy.dx = cot x. dx Dividing both sides by (1 + x2) 𝑑𝑦/𝑑π‘₯ + 2π‘₯𝑦/((1 + π‘₯2)) 𝑑π‘₯/𝑑π‘₯ = (cot⁑(π‘₯))/((1 + π‘₯^2)). 𝑑π‘₯/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ + 2π‘₯/((1 + π‘₯2)) y = cot⁑π‘₯/((1 + π‘₯2)) Step 2 : Find P and Q Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q where P = 2π‘₯/((1 + π‘₯^2)) & Q = cot⁑π‘₯/((1 + π‘₯^2)) Step 3 : Find Integrating factor, I.F I.F. = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— I.F. = e^∫1β–’γ€–2π‘₯/((1 + π‘₯^2 ) ) 𝑑π‘₯ γ€— Let t = 1 + x2 dt = 2x dx I.F. = e^∫1▒〖𝑑𝑑/𝑑 γ€— = elog t = t = (1 + x2) So, I.F = 1 + x2 Step 4 : Solution of the equation y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹.𝑑π‘₯+𝐢〗 Putting values, y.(1 + x2) = ∫1β–’cot⁑π‘₯/((1 + π‘₯2)) Γ— (1+π‘₯2).dx + c y.(1 + x2) = ∫1β–’γ€–cot⁑π‘₯ 𝑑π‘₯γ€—+𝑐 y (1 + x2) = log|sin⁑π‘₯ | + C Dividing by (1 + x2) y = (1 + x2)βˆ’1 log |𝐬𝐒𝐧⁑𝒙 |+𝒄(𝟏+"x2" )^(βˆ’πŸ) is the general solution of the given equation

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