Ex 9.5

Chapter 9 Class 12 Differential Equations (Term 2)
Serial order wise    ### 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﷯ 