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Ex 9.5, 15 Class 12 - Find solution 2xy + y^2 - 2x^2 dy/dx = 0, when

Ex 9.5, 15 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 15 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 15 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.5, 15 - Chapter 9 Class 12 Differential Equations - Part 5


Transcript

Ex 9.5, 15 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 2𝑥𝑦+𝑦^2−2𝑥^2 𝑑𝑦/𝑑𝑥=0;𝑦=2 When 𝑥=1 Differential equation can be written 𝑎s 2𝑥𝑦+𝑦^2−2𝑥^2 𝑑𝑦/𝑑𝑥=0 2𝑥𝑦+𝑦^2= 2𝑥^2 𝑑𝑦/𝑑𝑥 2𝑥^2 𝑑𝑦/𝑑𝑥=2𝑥𝑦+𝑦^2 𝑑𝑦/𝑑𝑥= (2𝑥𝑦 + 𝑦^2)/(2𝑥^2 ) 𝑑𝑦/𝑑𝑥= 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) …(1) Let F(x, y) = 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = 𝜆𝑦/𝜆𝑥 + 〖(𝜆𝑦)〗^2/(2〖(𝜆𝑥)〗^(2 ) ) = 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) = 𝜆° F(x, y) ∴ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x 𝑑𝑦/𝑑𝑥 = x 𝑑𝑣/𝑑𝑥 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥= 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) 𝑣+𝑥 𝑑𝑣/𝑑𝑥 = 𝑣𝑥/𝑥 + 1/2 (𝑣^2 𝑥^2)/𝑥^2 𝑣+𝑥 𝑑𝑣/𝑑𝑥 = 𝑣+ 𝑣^2/2 𝑑𝑦/𝑑𝑥= 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) 𝑣+𝑥 𝑑𝑣/𝑑𝑥 = 𝑣+ 𝑣^2/2 𝑥𝑑𝑣/𝑑𝑥 = 𝑣^2/2 2𝑑𝑣/𝑣^2 = 𝑑𝑥/𝑥 Integrating both sides 2∫1▒𝑑𝑣/𝑣^2 "=" ∫1▒𝑑𝑥/𝑥 2∫1▒〖𝑣^(−2) 𝑑𝑣=log⁡|𝑥|+𝑐〗 2 (𝑣^(−2 + 1) )/(−2 + 1) =log⁡|𝑥|+𝑐 2 (〖𝑣 〗^(−1) )/(−1) =log⁡|𝑥|+𝑐 (−2 )/𝑣 =log⁡|𝑥|+𝑐 2 (〖𝑣 〗^(−1) )/(−1) =log⁡|𝑥|+𝑐 (−2 )/𝑣 =log⁡|𝑥|+𝑐 Putting value of v = (𝑦 )/𝑥 (−2𝑥)/𝑦 = log |𝑥| + C Now, Putting x = 1 & y = 2 in (2) (−2(1))/2 = log |1| + C −1 = 0 + C C = −1 …(2) (As log 1 = 0) Putting value of C in (2) (−2𝑥)/𝑦 = log |𝑥| + C (−2𝑥)/𝑦 = log |𝑥| − 1 y = (−2𝑥)/(〖log 〗⁡|𝑥| − 1) y = 𝟐𝒙/〖𝟏 − 𝐥𝐨𝐠 〗⁡|𝒙|" "

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.