Ex 9.5, 15 - Find particular solution: 2xy + y2 - 2x2 dy/dx = 0 - Ex 9.5

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  1. Class 12
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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﷯−2 𝑥﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯=0;𝑦=2 When 𝑥=1 Differential equation can be written 𝑎s 2𝑥𝑦+ 𝑦﷮2﷯−2 𝑥﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯=0 2 𝑥﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯=2𝑥𝑦+ 𝑦﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯= 2𝑥𝑦 + 𝑦﷮2﷯﷮2 𝑥﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑦﷮𝑥﷯ + 𝑦﷮2﷯﷮2 𝑥﷮2﷯﷯ Let F(x, y) = 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯ + 𝑦﷮2﷯﷮2 𝑥﷮2﷯﷯ Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = 𝜆𝑦﷮𝜆𝑥﷯ + (𝜆𝑦)﷮2﷯﷮2 (𝜆𝑥)﷮2 ﷯﷯ = 𝑦﷮𝑥﷯ + 𝑦﷮2﷯﷮2 𝑥﷮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﷯﷮2 𝑥﷮2﷯﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣𝑥﷮𝑥﷯ + 1﷮2﷯ 𝑣﷮2﷯ 𝑥﷮2﷯﷮ 𝑥﷮2﷯﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣+ 𝑣﷮2﷯﷮2﷯ 𝑥𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 + 𝑣﷮2﷯﷮2﷯ − v 𝑥𝑑𝑣﷮𝑑𝑥﷯ = 𝑣﷮2﷯﷮2﷯ 2𝑑𝑣﷮ 𝑣﷮2﷯﷯ = 𝑑𝑥﷮𝑥﷯ Integrating both sides 2 ﷮﷮ 𝑑𝑣﷮ 𝑣﷮2﷯﷯ = 𝑓 𝑑𝑥﷮𝑥﷯﷯ 2 ﷮﷮ 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 Putting value of v = 𝑦 ﷮𝑥﷯ −2𝑥﷮𝑦﷯ = log 𝑥﷯ + C Put x = 1 & y = 2 −2(1)﷮2﷯ = log 1 + C − 1 = 0 + C C = −1 Putting value in (2) −2𝑥﷮𝑦﷯ = log 𝑥﷯ − 1 y = −2𝑥﷮ log ﷮ 𝑥﷯ − 1 ﷯﷯ y = 𝟐𝒙﷮ 𝟏 − 𝐥𝐨𝐠 ﷮ 𝒙﷯ ﷯﷯

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