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Misc 5 - Family of circles in first quadrant which touch axes - Formation of Differntial equation when general solution given

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Misc 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes . Drawing figure : Let C be the family of circles in first quadrant touching coordinate. Let radius be 𝑎 ∴ Center of circle = (−𝑎, 𝑎) Thus, Equation of a circle is 𝑥−𝑎﷯﷮2﷯+ 𝑦−𝑎﷯﷮2﷯= 𝑎﷮2﷯ 𝑥﷮2﷯+ 𝑎﷮2﷯−2𝑎𝑥+ 𝑦﷮2﷯+ 𝑎﷮2﷯−2𝑎𝑥= 𝑎﷮2﷯ 𝑥﷮2﷯+ 𝑦﷮2﷯−2𝑎𝑥−2𝑎𝑦+ 𝑎﷮2﷯=0 Since there is only one variable , we differentiate once Differentiating w.r.t. 𝑥 both Sides 𝑑 𝑥﷮2﷯ + 𝑦﷮2﷯ − 2𝑎𝑥 − 2𝑎𝑦+ 𝑎﷮2﷯﷯﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 0﷯ 2𝑥+2𝑦 𝑑𝑦﷮𝑑𝑥﷯−2𝑎−2𝑎 𝑑𝑦﷮𝑑𝑥﷯+0=0 𝑥+𝑦 𝑑𝑦﷮𝑑𝑥﷯=𝑎+ 𝑎𝑑𝑦﷮𝑑𝑥﷯ 𝑥+𝑦 𝑑𝑦﷮𝑑𝑥﷯=𝑎 1+ 𝑑𝑦﷮𝑑𝑥﷯﷯ a = 𝑥 + 𝑦𝑑𝑦﷮𝑑𝑥﷯﷮ 1 + 𝑑𝑦﷮𝑑𝑥﷯﷯﷯ a = 𝑥 + 𝑦 𝑦﷮′﷯﷮1 + 𝑦﷮′﷯﷯ Putting Value of 𝑎 in (1) 𝑥−𝑎﷯﷮2﷯+ 𝑦−𝑎﷯﷮2﷯= 𝑎﷮2﷯ 𝑥− 𝑥 + 𝑦 𝑦﷮′﷯﷮1 + 𝑦﷮′﷯﷯﷯﷯﷮2﷯+ 𝑦− 𝑥 + 𝑦 𝑦﷮′﷯﷮1 + 𝑦﷮′﷯﷯﷯﷯﷮2﷯= 𝑥 + 𝑦 𝑦﷮′﷯﷮1 + 𝑦﷮′﷯﷯﷯﷮2﷯ 𝑥 1 + 𝑦﷮′﷯﷯− 𝑥 + 𝑦 𝑦﷮′﷯﷯﷯﷮2﷯﷮ 1 + 𝑦﷮′﷯﷯﷮2﷯﷯+ 𝑦 1 + 𝑦﷮′﷯﷯− 𝑥 + 𝑦 𝑦﷮′﷯﷯﷯﷮2﷯﷮ 1 + 𝑦﷮′﷯﷯﷮2﷯﷯= 𝑥 + 𝑦 𝑦﷮′﷯﷯﷮2﷯﷮ 1 + 𝑦﷮′﷯﷯﷮2﷯﷯ 𝑥+𝑥 𝑦﷮′﷯−𝑥−𝑦 𝑦﷮′﷯﷯﷮2﷯+ 𝑦+𝑦 𝑦﷮′﷯−𝑥−𝑦 𝑦﷮′﷯﷯﷮2﷯= 𝑥+𝑦 𝑦﷮′﷯﷯﷮2﷯ 𝑥 𝑦﷮′﷯−𝑦 𝑦﷮′﷯﷯﷮2﷯+ 𝑦−𝑥﷯﷮2﷯= 𝑥+𝑦 𝑦﷮′﷯﷯﷮2﷯ 𝑥−𝑦﷯﷮2﷯ 𝑦﷮′﷯﷯﷮2﷯+ 𝑥−𝑦﷯﷮2﷯= 𝑥+𝑦 𝑦﷮′﷯﷯﷮2﷯ 𝑥−𝑦﷯﷮2﷯ 1+ 𝑦﷮′﷯﷯﷮2﷯﷯= 𝑥+𝑦 𝑦﷮′﷯﷯﷮2﷯ 𝒙+𝒚 𝒚﷮′﷯﷯﷮𝟐﷯= 𝒙−𝒚﷯﷮𝟐﷯ 𝟏+ 𝒚﷮′﷯﷯﷮𝟐﷯﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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