Example 25 - Family of circles in second quadrant, touching - Examples

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Example 25 - Chapter 9 Class 12 Differential Equations - Part 2

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Example 25 - Chapter 9 Class 12 Differential Equations - Part 3 Example 25 - Chapter 9 Class 12 Differential Equations - Part 4

  1. Chapter 9 Class 12 Differential Equations (Term 2)
  2. Serial order wise

Transcript

Example 25 From the differential equation of the family of circles in the second quadrant and touching the coordinate axes . Drawing figure : Let C be the family of circles in second quadrant touching coordinate. Let radius be 𝑎 ∴ Center of circle = (−𝑎, 𝑎) Equation representing family C is x−(−𝑎)﷯﷮2﷯+ 𝑦−𝑎﷯﷮2﷯= 𝑎﷮2﷯ x + 𝑎﷯﷮2﷯+ 𝑦−𝑎﷯﷮2﷯= 𝑎﷮2﷯ 𝑥2 + 𝑎2 + 2ax + y2 + 𝑎2 − 2𝑎y = 𝑎2 𝑥2 + 𝑦2 + 2ax − 2ay + 2𝑎2 = 𝑎2 𝑥2 + y2 + 2𝑎x − 2𝑎y + 𝑎2 = 0 Differentiate w.r.t x 2x + 2y. 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑎 − 2a 𝑑𝑦﷮𝑑𝑥﷯ + 0 = 0 x + y. 𝑑𝑦﷮𝑑𝑥﷯ + 𝑎 − 𝑎𝑑𝑦﷮𝑑𝑥﷯ = 0 x + y. 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑎 + 𝑎𝑑𝑦﷮𝑑𝑥﷯ x + y 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 𝑑𝑦﷮𝑑𝑥﷯ −1﷯ 𝑎 = 𝑥 + 𝑦 𝑑𝑦﷮𝑑𝑥﷯ ﷮ 𝑑𝑦﷮𝑑𝑥﷯ − 1﷯ 𝑎 = 𝒙 + 𝒚 𝒚﷮′﷯ ﷮ 𝒚﷮′﷯ − 𝟏﷯ Putting value of a in (1) x−(−𝑎)﷯﷮2﷯+ 𝑦−𝑎﷯﷮2﷯= 𝑎﷮2﷯ x− − 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷯﷮2﷯+ 𝑦− 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯ x+ 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯+ 𝑦− 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯ 𝑥 𝑦﷮′﷯− 1﷯ + 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯+ 𝑦 𝑦﷮′﷯− 1﷯ − 𝑥 − 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯ 𝑥 𝑦﷮′﷯ − 𝑥 + 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯+ −𝑥 − 𝑦 ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯ (𝑥 + 𝑦) 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯+ −(𝑥 + 𝑦) ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯ ﷮ 𝑦﷮′﷯ − 1﷯﷯﷮2﷯ (𝑥 + 𝑦) ﷮2﷯ ( 𝑦﷮′﷯)﷮2﷯+ (𝑥 + 𝑦) ﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯﷯﷮2﷯ (𝒙 + 𝒚) ﷮𝟐﷯ ( 𝒚﷮′﷯)﷮𝟐﷯ + 𝟏﷯= 𝒙 + 𝒚 𝒚﷮′﷯﷯﷮𝟐﷯ which is the required differential equation

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.