# Example 25

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 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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About the Author

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.