Ex 9.5, 14 - Find particular solution: dy/dx - y/x + cosec = 0 - Ex 9.5

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.5, 14 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑑𝑦/𝑑π‘₯βˆ’π‘¦/π‘₯+π‘π‘œπ‘ π‘’π‘(𝑦/π‘₯)=0;𝑦=0 When π‘₯=1 Differential equation is 𝑑𝑦/𝑑π‘₯ = 𝑦/π‘₯βˆ’π‘π‘œπ‘ π‘’π‘(𝑦/π‘₯) Let F(x, y) = 𝑑𝑦/𝑑π‘₯ = 𝑦/π‘₯βˆ’π‘π‘œπ‘ π‘’π‘(𝑦/π‘₯) Finding F(πœ†x, πœ†y) F(πœ†x, πœ†y) = ("πœ†" 𝑦)/("πœ†" π‘₯)βˆ’π‘π‘œπ‘ π‘’π‘(("πœ†" 𝑦)/("πœ†" π‘₯)) = 𝑦/π‘₯ βˆ’ cosec (𝑦/π‘₯) = πœ†Β° F(x, y) ∴ F(x, y) is π‘Ž homogenous function of degree zero F(πœ†x, πœ†y) = πœ†Β° F(x , y) Putting y = vx Diff w.r.t. x 𝑑𝑦/𝑑π‘₯ = x 𝑑𝑣/𝑑π‘₯ + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯ = 𝑦/π‘₯βˆ’π‘π‘œπ‘ π‘’π‘(𝑦/π‘₯) v + x 𝑑𝑣/𝑑π‘₯ = 𝑣π‘₯/π‘₯ βˆ’ cosec (𝑣π‘₯/π‘₯) v + x 𝑑𝑣/𝑑π‘₯ = v βˆ’ cosec v (π‘₯ 𝑑𝑣)/𝑑π‘₯ = v βˆ’ cosec v βˆ’ v (π‘₯ 𝑑𝑣)/𝑑π‘₯ = βˆ’ cosec v (βˆ’π‘‘π‘£)/(π‘π‘œπ‘ π‘’π‘ 𝑣) = 𝑑π‘₯/π‘₯ Integrating both sides ∫1β–’γ€–(βˆ’π‘‘π‘£)/(π‘π‘œπ‘ π‘’π‘ 𝑣) " = " ∫1▒𝑑π‘₯/π‘₯γ€— ∫1β–’γ€–βˆ’sin⁑𝑣 𝑑𝑣〗=log⁑〖|π‘₯|+𝑐〗 Put value of v = 𝑦/π‘₯ cos 𝑦/π‘₯ = log |π‘₯| + C Putting x = 1 & y = 0 cos 0/1 = log 1 + C 1 = 0 + C C = 1 Putting value in (2) cos 𝑦/2 = log |π‘₯| + 1 cos 𝑦/π‘₯ = log |π‘₯| + log e cos π’š/𝒙 = log |𝒆𝒙|

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