Example 18

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