# Example 25 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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−𝑎2= 𝑎2 x + 𝑎2+ 𝑦−𝑎2= 𝑎2 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. 𝑑𝑦𝑑𝑥 + 2𝑎 − 2a 𝑑𝑦𝑑𝑥 + 0 = 0 x + y. 𝑑𝑦𝑑𝑥 + 𝑎 − 𝑎𝑑𝑦𝑑𝑥 = 0 x + y. 𝑑𝑦𝑑𝑥 = − 𝑎 + 𝑎𝑑𝑦𝑑𝑥 x + y 𝑑𝑦𝑑𝑥 = 𝑎 𝑑𝑦𝑑𝑥 −1 𝑎 = 𝑥 + 𝑦 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 − 1 𝑎 = 𝒙 + 𝒚 𝒚′ 𝒚′ − 𝟏 Putting value of a in (1) x−(−𝑎)2+ 𝑦−𝑎2= 𝑎2 x− − 𝑥 + 𝑦 𝑦′ 𝑦′ − 12+ 𝑦− 𝑥 + 𝑦 𝑦′ 𝑦′ − 12= 𝑥 + 𝑦 𝑦′ 𝑦′ − 12 x+ 𝑥 + 𝑦 𝑦′ 𝑦′ − 12+ 𝑦− 𝑥 + 𝑦 𝑦′ 𝑦′ − 12= 𝑥 + 𝑦 𝑦′ 𝑦′ − 12 𝑥 𝑦′− 1 + 𝑥 + 𝑦 𝑦′ 𝑦′ − 12+ 𝑦 𝑦′− 1 − 𝑥 − 𝑦 𝑦′ 𝑦′ − 12= 𝑥 + 𝑦 𝑦′ 𝑦′ − 12 𝑥 𝑦′ − 𝑥 + 𝑥 + 𝑦 𝑦′ 𝑦′ − 12+ −𝑥 − 𝑦 𝑦′ − 12= 𝑥 + 𝑦 𝑦′ 𝑦′ − 12 (𝑥 + 𝑦) 𝑦′ 𝑦′ − 12+ −(𝑥 + 𝑦) 𝑦′ − 12= 𝑥 + 𝑦 𝑦′ 𝑦′ − 12 (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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Example 25 Important You are here

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Chapter 9 Class 12 Differential Equations

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.