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Example 15 - Show (x - y) dy/dx = x + 2y is homogeneous, solve - Solving homogeneous differential equation

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Example 15 Show that the differential equation 𝑥−𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯=𝑥+2𝑦 is homogeneous and solve it. Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ 𝑥−𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯=𝑥+2𝑦 𝑑𝑦﷮𝑑𝑥﷯= 𝑥 + 2𝑦﷮𝑥 − 𝑦﷯﷯ Step 2: Put F 𝑥 , 𝑦﷯= 𝑑𝑦﷮𝑑𝑥﷯ & Find F 𝜆𝑥 ,𝜆𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑥 + 2𝑦﷮𝑥 − 𝑦﷯﷯ Put F 𝑥 , 𝑦﷯= 𝑥 + 2𝑦﷮𝑥 − 𝑦﷯﷯ Finding F 𝜆𝑥 ,𝜆𝑦﷯ F 𝜆𝑥 ,𝜆𝑦﷯= 𝜆𝑥 + 2 𝜆𝑦﷯﷮𝜆𝑥 −𝜆𝑦﷯ = 𝜆 𝑥 + 2𝑦﷯﷮𝜆 𝑥 − 𝑦﷯﷯ = 𝑥 + 2𝑦﷯﷮𝑥 − 𝑦﷯ = F 𝑥 , 𝑦﷯ Thus , F 𝜆𝑥 ,𝜆𝑦﷯=F 𝑥 , 𝑦﷯ =𝜆° F 𝑥 , 𝑦﷯ Thus , F 𝑥 , 𝑦﷯ is Homogeneous function of degree zero Therefore, the given Differential Equation is Homogeneous differential Equation Step 3: Solving 𝑑𝑦﷮𝑑𝑥﷯ by Putting 𝑦=𝑣𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑥 + 2𝑦﷮𝑥 − 𝑦﷯﷯ Let 𝑦=𝑣𝑥 So , 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑣𝑥﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑𝑣﷮𝑑𝑥﷯ . 𝑥+𝑣 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑𝑣﷮𝑑𝑥﷯ 𝑥+𝑣 Putting 𝑑𝑦﷮𝑑𝑥﷯ 𝑎nd 𝑦﷮𝑥﷯ 𝑖𝑛 𝑖﷯ 𝑖.𝑒. 𝑑𝑦﷮𝑑𝑥﷯= 𝑥 + 2𝑦﷮𝑥 − 𝑦﷯ 𝑑𝑣﷮𝑑𝑥﷯ 𝑥+𝑣= 𝑥 + 2 𝑣𝑥﷮𝑥 − 𝑣𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ 𝑥+𝑣= 𝑥 1 +2𝑣﷯﷮𝑥 1 − 𝑣﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ 𝑥+𝑣= 1+2𝑣﷮1− 𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯ 𝑥= 1+2𝑣﷮1− 𝑣﷯−𝑣 𝑑𝑣﷮𝑑𝑥﷯ . 𝑥= 1 + 2𝑣 − 𝑣 1− 𝑣﷯﷮1 −𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯ . 𝑥= 1 + 2𝑣 − 𝑣 + 𝑣﷮2﷯﷮1 − 𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯ . 𝑥= 𝑣﷮2﷯ + 𝑣+1﷮1 − 𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯ 𝑥=− 𝑣﷮2﷯ + 𝑣 + 1﷮𝑣 − 1﷯﷯ 𝑑𝑣 𝑣−1﷮ 𝑣﷮2 ﷯+ 𝑣 + 1﷯﷯= − 𝑑𝑥﷮𝑥﷯ Integrating Both Sides ﷮﷮ 𝑣 −1﷮ 𝑣﷮2﷯ + 𝑣 + 1﷯𝑑𝑣= ﷮﷮ −𝑑𝑥﷮𝑥﷯﷯﷯ ﷮﷮ 𝑣 −1﷮ 𝑣﷮2﷯ + 𝑣 + 1﷯𝑑𝑣=− ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯﷯ ﷮﷮ 𝑣 −1﷯ 𝑑𝑣﷮ 𝑣﷮2﷯ + 𝑣 + 1﷯𝑑𝑣﷯=− log﷮|𝑥|﷯ + 𝑐 Putting 𝑣﷮2﷯+𝑣+1= 𝑣+ 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯ and 𝑣−1=𝑣+ 𝟏﷮𝟐﷯− 𝟏﷮𝟐﷯−1 = 𝑣+ 1﷮2﷯﷯− 3﷮2﷯ ﷮﷮ 𝑣 + 1﷮2﷯﷯ − 3﷮2﷯﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷯ 𝑑𝑣=− log﷮𝑥+𝑐﷯ ﷮﷮ 𝑣 + 1﷮2﷯﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷯𝑑𝑣− 3﷮2﷯ ﷮﷮ 1﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷯𝑑𝑣=− log﷮𝑥+𝑐﷯ Thus, I = I1 – 3 ﷮2﷯ I2 Solving 𝐼1 𝐼1= ﷮﷮ 𝑣 + 1﷮2﷯﷯﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷯𝑑𝑣 Put 𝑣+ 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯ =𝑡 Diff. w.r.t. 𝑣 𝑑 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷮𝑑𝑣﷯= 𝑑𝑡﷮𝑑𝑣﷯ 2 𝑣+ 1﷮2﷯﷯= 𝑑𝑡﷮𝑑𝑣﷯ 𝑑𝑣= 𝑑𝑡﷮2 𝑣 + 1﷮2﷯﷯﷯ Putting value of v & dv in I1 𝐼1= ﷮﷮ 𝑣 + 1﷮2﷯﷯﷮𝑡﷯﷯ × 𝑑𝑡﷮2 𝑣 + 1﷮2﷯﷯﷯ = 1﷮2﷯ ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ = 1﷮2﷯ log﷮ ﷯ 𝑡﷯ Putting 𝑡= 𝑣+ 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯ = 1﷮2﷯𝑙𝑜𝑔 𝑣+ 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯ = 1﷮2﷯𝑙𝑜𝑔 𝑣﷮2﷯+2𝑣 × 1﷮2﷯ + 1﷮4﷯ + 3﷮4﷯﷯ I1 = 1﷮2﷯ log﷮ 𝑣﷮2﷯+𝑣+1﷯﷯ Solving 𝑰𝟐 𝐼2= ﷮﷮ 𝑑𝑣﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯﷯ = ﷮﷮ 𝑑𝑣﷮ 𝑣 + 1﷮2﷯﷯﷮2﷯+ ﷮3﷯﷮2﷯﷯﷮2﷯﷯﷯ Put 𝑡=𝑣+ 1﷮2﷯ Diff. w.r.t. 𝑣 𝑑𝑡﷮𝑑𝑣﷯=1 ⇒ 𝑑𝑡=𝑑𝑣 = ﷮﷮ 𝑑𝑡﷮ 𝑡﷮2﷯ + ﷮3﷯﷮2﷯﷯﷮2﷯﷯﷯ = 1﷮ ﷮3﷯﷮2﷯﷯ tan﷮−1﷯﷮ 𝑡﷮ ﷮3﷯﷮2﷯﷯﷯ = 2﷮ ﷮3﷯﷯ tan﷮−1﷯﷮ 2 𝑣 + 1﷮2﷯﷯﷮ ﷮3﷯﷯﷯ = 2﷮ ﷮3﷯﷯ tan﷮−1﷯﷮ 2𝑣 + 1﷮ ﷮3﷯﷯﷯﷯ Hence I = 𝐼1− 3﷮2﷯ 𝐼2 I = 1﷮2﷯ log﷮ ﷯ 𝑣﷮2﷯+𝑣+1﷯− 3﷮2﷯ × 2﷮ ﷮3﷯﷯ tan﷮−1﷯﷮ 2𝑣 +1﷮ ﷮3﷯﷯﷯﷯ I = 1﷮2﷯ log﷮ ﷯ 𝑣﷮2﷯+𝑣+1﷯− ﷮3﷯ tan﷮−1﷯﷮ 2𝑣 +1﷮ ﷮3﷯﷯﷯﷯ Replacing v by 𝑦 ﷮𝑥﷯ I = 1﷮2﷯𝑙𝑜𝑔 𝑦﷮𝑥﷯﷯﷮2﷯+ 𝑦﷮𝑥﷯+1﷯− ﷮3﷯ tan﷮−1﷯﷮ 2𝑦﷮𝑥﷯ + 1﷮ ﷮3﷯﷯﷯﷯ I = 1﷮2﷯𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯− ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯ Putting Value of I in (2) 1﷮2﷯𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯− ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯=−𝑙𝑜𝑔 𝑥﷯+𝑐 1﷮2﷯𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯+𝑙𝑜𝑔 𝑥﷯= ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯+𝑐 Multiplying Both Sides By 2 𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯+2 𝑙𝑜𝑔 𝑥﷯=2 ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯+2𝑐 𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯+𝑙𝑜𝑔 𝑥﷯﷮2﷯=2 ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯+2𝑐 Put 2𝑐=𝑐 𝑙𝑜𝑔 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑦﷮𝑥﷯+1﷯ × 𝑥﷮2﷯﷯﷯=2 ﷮3﷯ tan﷮−1﷯﷮ 2𝑦 + 𝑥﷮ ﷮3﷯ 𝑥﷯﷯﷯+𝑐 𝑙𝑜𝑔 𝑥﷮2﷯𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑥﷮2﷯𝑦﷮𝑥﷯+ 𝑥﷮2﷯﷯=2 ﷮3﷯ tan﷮−1﷯﷮ 𝑥 + 2𝑦﷮ ﷮3﷯ 𝑥﷯﷯﷯+𝑐 𝒍𝒐𝒈 𝒙﷮𝟐﷯+𝒙𝒚+ 𝒚﷮𝟐﷯﷯=𝟐 ﷮𝟑﷯ 𝒕𝒂𝒏﷮−𝟏﷯﷮ 𝒙 + 𝟐𝒚﷮ ﷮𝟑﷯ 𝒙﷯﷯﷯+𝒄 Is the General Solution of the Differential Equation given

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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.
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