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


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Ex 9.4, 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(𝝀x, 𝝀y) 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 π’…π’š/𝒅𝒙 = x 𝒅𝒗/𝒅𝒙 + 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/𝒅𝒙 = 𝒗𝒙/𝒙 βˆ’ cosec (𝒗𝒙/𝒙) 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’—)/(𝒄𝒐𝒔𝒆𝒄 𝒗) = 𝒅𝒙/𝒙 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 π’š/𝒙 = log |𝒙| + 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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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.