Ex 9.6, 19 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.6, 19 The integrating Factor of the differential equation 1− 𝑦﷮2﷯﷯ 𝑑𝑥﷮𝑑𝑦﷯+𝑦𝑥=𝑎𝑦 −1<𝑦<1﷯ is (A) 1﷮ 𝑦﷮2﷯−1﷯ (B) 1﷮ ﷮ 𝑦﷮2﷯−1﷯﷯ (C) 1﷮1− 𝑦﷮2﷯﷯ (D) 1﷮ ﷮1− 𝑦﷮2﷯﷯﷯ 1− 𝑦﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯+𝑦𝑥=𝑎𝑦 Dividing both sides by 1 − y2 𝑑𝑦﷮𝑑𝑥﷯ + 𝑦𝑥﷮1− 𝑦﷮2﷯﷯ = 𝑎𝑦﷮1− 𝑦﷮2﷯﷯ Differential equation is of the form 𝑑𝑦﷮𝑑𝑥﷯ + P1x = Q1 where P1 = 𝑦﷮1 − 𝑦﷮2﷯﷯ & Q1 = 𝑎𝑦﷮1 − 𝑦﷮2﷯﷯ IF = 𝑒﷮ ﷮﷮𝑝1𝑑𝑥﷯﷯ Finding ﷮﷮𝑷𝟏 𝒅𝒚﷯ ﷮﷮𝑃1 𝑑𝑦=﷯ ﷮﷮ 𝑦﷮1− 𝑦﷮2﷯﷯ 𝑑𝑦 ﷯ Putting 1 − y2 = t −2y dy = dt y dy = −1﷮2﷯ dt ∴ Our equation becomes ﷮﷮𝑃1 𝑑𝑦= −1﷮2﷯ ﷯ ﷮﷮ 𝑑𝑡﷮𝑡﷯ ﷯ ﷮﷮𝑃1 𝑑𝑦= −1﷮2﷯ ﷯ log﷮𝑡﷯ Putting back value of t ﷮﷮𝑃1 𝑑𝑦= −1﷮2﷯ ﷯ log⁡(1− 𝑦﷮2﷯) ﷮﷮𝑃1 𝑑𝑦=﷯ log﷮ 1− 𝑦﷮2﷯﷯﷯﷮ −1﷮2﷯﷯ ﷮﷮𝑃1 𝑑𝑦= ﷯log 1﷮ ﷮1 − 𝑦﷮2﷯﷯﷯ Thus, IF = 𝑒﷮ ﷮﷮𝑝1𝑑𝑥﷯﷯ IF = e﷮log 1﷮ ﷮1 − 𝑦﷮2﷯﷯﷯﷯ IF = 1﷮ ﷮1 − 𝑦﷮2﷯﷯﷯ ∴ Part (D) is correct answer.