Check sibling questions

Ex 9.5, 13 - Find particular solution: [x sin2 (y/x) - y] dx - Solving homogeneous differential equation

Ex 9.5, 13 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 13 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 13 - Chapter 9 Class 12 Differential Equations - Part 4

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Transcript

Ex 9.5, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑥 sin﷮2﷯﷮ 𝑦﷮𝑥﷯﷯−𝑦﷯﷯𝑑𝑥+𝑥𝑑𝑦=0;𝑦= 𝜋﷮4﷯ When 𝑥=1 Differential equation can be written as 𝑥 sin﷮2﷯﷮ 𝑦﷮𝑥﷯﷯−𝑦﷯﷯𝑑𝑥+𝑥𝑑𝑦=0 x dy = − 𝑥 sin﷮2﷯﷮ 𝑦﷮𝑥﷯﷯−𝑦﷯﷯dx 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑥﷮𝑥﷯ sin﷮2﷯﷮ 𝑦﷮𝑥﷯ − 𝑦﷮𝑥﷯﷯﷯ Let F(x, y) = 𝑑𝑦﷮𝑑𝑥﷯ =− 𝑥﷮𝑥﷯ sin﷮2﷯﷮ 𝑦﷮𝑥﷯ − 𝑦﷮𝑥﷯﷯﷯ Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = − sin﷮2﷯﷮ 𝜆𝑦﷮𝜆𝑥﷯− 𝜆𝑦﷮𝜆𝑥﷯﷯﷯ = − sin﷮2﷯﷮ 𝑦﷮𝑥﷯− 𝑦﷮𝑥﷯﷯﷯ = 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﷯ = − 𝑥﷮𝑥﷯ sin﷮2﷯﷮ 𝑦﷮𝑥﷯ − 𝑦﷮𝑥﷯﷯﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = − sin﷮2﷯ 𝑣𝑥﷮𝑥﷯− 𝑣𝑥﷮𝑥﷯﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = − sin﷮2﷯𝑣−𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = v − sin2 v − v 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = − sin2 v 𝑑𝑣﷮ sin﷮2﷯﷮𝑣﷯﷯ = − 𝑑𝑥﷮𝑥﷯ Integrating both sides ﷮﷮ 𝑑𝑣﷮ sin﷮2﷯𝑣﷯﷯ = − ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮𝑐𝑜𝑠𝑒 𝑐﷮2﷯ 𝑣 𝑑𝑣﷯ = − ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ − cot v = − log 𝑥﷯+𝑐 log 𝑥﷯ − cot v = C Putting value of v = 𝑦﷮𝑥﷯ log 𝑥﷯ − cot 𝑦﷮𝑥﷯﷯ = C Putting x = 1 & y = 𝜋﷮4﷯ log 1 − cot 𝜋﷮4﷯﷯ = C 0 − 1 = C C = −1 Putting value of C in (2) log 𝑥﷯ − cot 𝑦﷮𝑥﷯ = −1 cot 𝑦﷮𝑥﷯ − log 𝑥﷯ = 1 cot 𝑦﷮𝑥﷯ − log 𝑥﷯ = log e cot 𝑦﷮𝑥﷯ = log 𝑥﷯ + log e cot 𝒚﷮𝒙﷯ = log 𝒆𝒙﷯

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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.