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

Transcript

Misc 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes . Let’s first draw the figure Let C be the family of circles in first quadrant touching coordinate axes Let radius be π‘Ž ∴ Center of circle = (βˆ’π‘Ž, π‘Ž) Thus, Equation of a circle is (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘Ž)^2=π‘Ž^2 π‘₯^2+π‘Ž^2βˆ’2π‘Žπ‘₯+𝑦^2+π‘Ž^2βˆ’2π‘Žπ‘¦=π‘Ž^2 π‘₯^2+𝑦^2βˆ’2π‘Žπ‘₯βˆ’2π‘Žπ‘¦+π‘Ž^2=0 Since there is only one variable, we differentiate once Differentiating w.r.t. π‘₯ (π‘₯^2 + 𝑦^2 βˆ’ 2π‘Žπ‘₯ βˆ’ 2π‘Žπ‘¦+π‘Ž^2 )^β€²=(0)^β€² 2π‘₯+2π‘¦π‘¦β€²βˆ’2π‘Žβˆ’2π‘Žπ‘¦β€²+0=0 π‘₯+π‘¦π‘¦β€²βˆ’π‘Žβˆ’π‘Žπ‘¦β€²=0 π‘₯+𝑦𝑦^β€²=π‘Ž+π‘Žπ‘¦^β€² π‘₯+𝑦𝑦^β€²=π‘Ž(1+𝑦^β€²) π‘Ž = (π‘₯ + 𝑦𝑦^β€²)/(1 + 𝑦^β€² ) Putting Value of π‘Ž in (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘Ž)^2=π‘Ž^2 (π‘₯βˆ’[(π‘₯ + 𝑦 𝑦^β€²)/(1 + 𝑦^β€² )])^2+(π‘¦βˆ’[(π‘₯ + 𝑦 𝑦^β€²)/(1 + 𝑦^β€² )])^2=((π‘₯ + 𝑦𝑦^β€²)/(1 + 𝑦^β€² ))^2 (π‘₯(1 + 𝑦^β€² ) βˆ’ (π‘₯ + 𝑦𝑦^β€² ))^2/(1 + 𝑦^β€² )^2 +(𝑦(1 + 𝑦^β€² ) βˆ’ (π‘₯ + 𝑦𝑦^β€² ))^2/(1 + 𝑦^β€² )^2 =(π‘₯ + 𝑦𝑦^β€² )^2/(1 + 𝑦^β€² )^2 (π‘₯+π‘₯𝑦^β€²βˆ’π‘₯βˆ’π‘¦π‘¦^β€² )^2+(𝑦+𝑦𝑦^β€²βˆ’π‘₯βˆ’π‘¦π‘¦^β€² )^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯𝑦^β€²βˆ’π‘¦π‘¦^β€² )^2+(π‘¦βˆ’π‘₯)^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯βˆ’π‘¦)^2 (𝑦^β€² )^2+(π‘₯βˆ’π‘¦)^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯βˆ’π‘¦)^2 (1+〖𝑦^β€²γ€—^2 )=(π‘₯+𝑦𝑦^β€² )^2 (𝒙+π’šπ’š^β€² )^𝟐=(π’™βˆ’π’š)^𝟐 (𝟏+γ€–π’š^β€²γ€—^𝟐 )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.