Example 18  Show curves, slope of tangent is x2 + y2 / 2xy - Solving homogeneous differential equation

Slide31.JPG
Slide32.JPG Slide33.JPG Slide34.JPG Slide35.JPG Slide36.JPG Slide37.JPG Slide38.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Example 18 Show that the family of curves for which the slope of the tangent at any point 𝑥 , 𝑦﷯ on its 𝑥﷮2﷯+ 𝑦﷮2﷯﷮2𝑥𝑦﷯ , is given by 𝑥﷮2﷯− 𝑦﷮2﷯=𝑐𝑥 We know that the slope of the tangent at 𝑥 ,𝑦﷯ of a curve is 𝑑𝑦﷮𝑑𝑥﷯ Given slope of tangent at 𝑥 , 𝑦﷯ is 𝑥﷮2﷯ + 𝑦﷮2﷯﷮2𝑥𝑦﷯ Therefore 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑦﷮2﷯﷮2𝑥𝑦﷯ Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑦﷮2﷯﷮2𝑥𝑦﷯ Step 2: Put 𝑑𝑦﷮𝑑𝑥﷯= F 𝑥 𝑦﷯ So, F 𝑥 𝑦﷯= 𝑥﷮2﷯ + 𝑦﷮2﷯﷮2𝑥𝑦﷯ Finding F 𝜆𝑥 ,𝜆𝑦﷯ F 𝜆𝑥 ,𝜆𝑦﷯= 𝜆𝑥﷯﷮2﷯ + 𝜆𝑦﷯﷮2﷯﷮2 𝜆𝑥﷯ 𝜆𝑦﷯﷯ = 𝜆﷮2﷯ 𝑥﷮2﷯ + 𝜆﷮2﷯ 𝑦﷮2﷯﷮2 𝜆﷮2﷯ 𝑥𝑦﷯ = 𝜆﷮2﷯ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯﷮ 𝜆﷮2﷯ 2𝑥𝑦﷯ = 𝑥﷮2﷯ + 𝑦﷮2﷯﷯﷮ 2𝑥𝑦﷯ = F 𝑥 , 𝑦﷯ So , F 𝜆𝑥 ,𝜆𝑦﷯= F 𝑥 , 𝑦﷯ =𝜆° F 𝑥 , 𝑦﷯ So , F 𝑥 , 𝑦﷯ is homogeneous function of degree zero, Therefore given equation is a homogeneous differential equation Step 3: Solving 𝑑𝑦﷮𝑑𝑥﷯ by putting 𝑦=𝑣𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑦﷮2﷯﷮2𝑥𝑦﷯ Put 𝑦=𝑣𝑥 Diff. w.r.t. 𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑣𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯=𝑥 𝑑𝑣﷮𝑑𝑥﷯+𝑣 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑥𝑑𝑣﷮𝑑𝑥﷯+𝑣 Putting values 𝑜𝑓 𝑑𝑦﷮𝑑𝑥﷯ and y in (i) v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + (𝑣 𝑥)﷮2﷯﷮2𝑥 (𝑣𝑥)﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑣﷮2﷯ 𝑥﷮2﷯﷮2 𝑥﷮2﷯ 𝑣﷯ − 𝑣 v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑣﷮2﷯ 𝑥﷮2﷯ − 2 𝑥﷮2﷯ 𝑣 . 𝑣﷮2 𝑥﷮2﷯ 𝑣﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑣﷮2﷯ 𝑥﷮2﷯ − 2 𝑥﷮2﷯ 𝑣﷮2﷯﷮2 𝑥﷮2﷯ 𝑣﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑣﷮2﷯ 𝑥﷮2﷯ − 2 𝑣﷮2﷯ 𝑥﷮2﷯﷮2 𝑥﷮2﷯ 𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ + 𝑣﷮2﷯ 𝑥﷮2﷯ − 2 𝑣﷮2﷯ 𝑥﷮2﷯﷮2 𝑥﷮2﷯ 𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ − 𝑣﷮2﷯ 𝑥﷮2﷯﷮2 𝑥﷮2﷯ 𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮2﷯ 1 − 𝑣﷮2﷯﷯﷮ 𝑥﷮2﷯. 2𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 1 − 𝑣﷮2﷯﷮2𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯= 1 − 𝑣﷮2﷯﷮2𝑣﷯ . 1﷮𝑥﷯ 2𝑣 𝑑𝑣﷮1 − 𝑣﷮2﷯﷯= 𝑑𝑥﷮𝑥﷯ 2𝑣 𝑑𝑣﷮− 𝑣﷮2﷯ − 1﷯﷯= 𝑑𝑥﷮𝑥﷯ 2𝑣 𝑑𝑣 ﷮ 𝑣﷮2﷯ − 1﷯= −𝑑𝑥﷮𝑥﷯ Integrating Both Sides ﷮﷮ 2𝑣 𝑑𝑣 ﷮ 𝑣﷮2﷯−1﷯= ﷮﷮ −𝑑𝑥﷮𝑥﷯﷯﷯ ﷮﷮ 2𝑣 𝑑𝑣 ﷮ 𝑣﷮2﷯ − 1﷯= ﷮﷮ −𝑑𝑥﷮𝑥﷯﷯﷯ ﷮﷮ 2𝑣 ﷮ 𝑣﷮2﷯ − 1﷯﷯𝑑𝑣=−𝑙𝑜𝑔 𝑥﷯+𝑐 Solving ﷮﷮ 𝟐𝒗 ﷮ 𝒗﷮𝟐﷯ − 𝟏﷯﷯𝒅𝒗 Put 𝑣﷮2﷯−1=𝑡 Diff. w.r.t. 𝑣 𝑑 𝑣﷮2﷯ −1﷯﷮𝑑𝑣﷯= 𝑑𝑡﷮𝑑𝑣﷯ 2𝑣= 𝑑𝑡﷮𝑑𝑣﷯ 𝑑𝑣= 𝑑𝑡﷮2𝑣﷯ ﷮﷮ 2𝑣﷮ 𝑣﷮2﷯ −1﷯﷯𝑑𝑣 = ﷮﷮ 2𝑣﷮𝑡﷯ ﷯ 𝑑𝑡﷮2𝑣﷯ = ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ = log 𝑡﷯ Putting t = v2 – 1 = log 𝑣﷮2﷯−1﷯ From (2) ﷮﷮ 2𝑣 ﷮ 𝑣﷮2﷯ − 1﷯﷯𝑑𝑣=−𝑙𝑜𝑔 𝑥﷯+𝑐 log 𝑣﷮2﷯−1﷯=−𝑙𝑜𝑔 𝑥﷯+𝑐1 Putting 𝑣𝑥=𝑦 or 𝑣= 𝑦﷮𝑥﷯ log 𝑦﷮𝑥﷯﷯﷮2﷯−1﷯=−𝑙𝑜𝑔 𝑥﷯+𝑐1 log 𝑦﷮𝑥﷯﷯﷮2﷯−1﷯=−𝑙𝑜𝑔 𝑥﷯+𝑐1 log 𝑦﷮𝑥﷯﷯﷮2﷯−1﷯+𝑙𝑜𝑔 𝑥﷯=+𝑐1 𝑙𝑜𝑔 𝑦﷮𝑥﷯﷯﷮2﷯−1﷯𝑥﷯=𝑐1 𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯ −1﷯𝑥﷯=𝑐1 Putting 𝑐1= log﷮𝑐﷯ 𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯−1﷯𝑥﷯= log﷮ ﷯𝑐1 Removing log 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯ −1﷯𝑥=𝐶1 𝑥 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯−𝑥=𝐶1 𝑦﷮2﷯﷮𝑥﷯−𝑥=𝐶1 𝑦﷮2﷯− 𝑥﷮2﷯﷮𝑥﷯=𝐶1 𝑦﷮2﷯− 𝑥﷮2﷯=𝐶1𝑥 𝑥﷮2﷯− 𝑦﷮2﷯=−𝐶1𝑥 Put 𝑐=−𝐶1 𝒙﷮𝟐﷯− 𝒚﷮𝟐﷯=𝒄𝒙 Hence Proved

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.