Misc 4 - Prove x2 - y2 = c(x2 + y2)2 is general solution of - Miscellaneous

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Misc 4 Prove that 𝑥﷮2﷯− 𝑦﷮2﷯=𝑐 𝑥﷮2﷯+ 𝑦﷮2﷯﷯﷮2﷯ is the general solution of differential equation 𝑥﷮3﷯−3𝑥 𝑦﷮2﷯﷯𝑑𝑥= 𝑦﷮3﷯−3 𝑥﷮2﷯𝑦﷯𝑑𝑦, where 𝑐 is a parameter . 𝑥﷮3﷯−3𝑥 𝑦﷮2﷯﷯𝑑𝑥= 𝑦﷮3﷯−3 𝑥﷮2﷯𝑦﷯𝑑𝑦 Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ 𝑥﷮3﷯−3𝑥 𝑦﷮2﷯﷯𝑑𝑥= 𝑦﷮3﷯−3 𝑥﷮2﷯𝑦﷯𝑑𝑦 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯= 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3 ﷯− 3 𝑥﷮2﷯𝑦﷯ Step 2: Put F 𝑥 , 𝑦﷯= 𝑑𝑦﷮𝑑𝑥﷯ and Find F 𝜆𝑥 ,𝜆𝑦﷯ F 𝑥 , 𝑦﷯= 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯ F 𝜆𝑥 ,𝜆𝑦﷯= 𝜆𝑥﷯﷮3﷯ − 3𝜆𝑥 . 𝜆𝑦﷯﷮2﷯﷮ 𝜆𝑦﷯﷮3﷯ − 3 𝜆﷮2﷯ 𝑥﷮2﷯ . 𝜆𝑦﷯ = 𝜆﷮3﷯ 𝑥﷮3﷯− 3𝜆 . 𝜆﷮2﷯ 𝑥 𝑦﷮2﷯﷮ 𝜆﷮3﷯ 𝑦﷮3﷯− 3𝜆 𝑥﷮2﷯ . 𝜆𝑦﷯ = 𝜆﷮3﷯ 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷯﷮ 𝜆﷮3﷯ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯﷯ = 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯ = F 𝑥 , 𝑦﷯ ∴ F 𝜆𝑥 ,𝜆𝑦﷯= F 𝑥 , 𝑦﷯ = 𝜆° F 𝑥 , 𝑦﷯ Hence F 𝑥 , 𝑦﷯ is a Homogeneous Differential Equation of degree O. So, 𝑑𝑦﷮𝑑𝑥﷯ is a Homogeneous Differential Equation Step 3: Solve 𝑑𝑦﷮𝑑𝑥﷯ by Putting 𝑦=𝑥𝑣 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯ Let y = x v Diff Both sides w.r.t. x 𝑑𝑦﷮𝑑𝑥﷯=𝑣 𝑑𝑥﷮𝑑𝑥﷯+𝑥 𝑑𝑣﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯=𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯ Putting Value of 𝑑𝑦﷮𝑑𝑥﷯ and y in (1) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑥﷮3﷯ − 3𝑥 𝑦﷮2﷯﷮ 𝑦﷮3﷯ − 3 𝑥﷮2﷯𝑦﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯= 𝑥﷮3﷯ − 3𝑥 𝑣𝑥﷯﷮2﷯﷮ 𝑣𝑥﷯﷮3﷯ − 3 𝑥﷮2﷯ 𝑣𝑥﷯﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥﷮3 ﷯− 3 𝑥﷮3﷯ . 𝑣﷮2﷯﷮ 𝑣﷮3﷯ 𝑥﷮3﷯ − 3 𝑥﷮3 . ﷯𝑣﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥﷮3﷯ 1 − 3 𝑣﷮2﷯﷯﷮ 𝑥﷮3﷯ 𝑣﷮3﷯ − 3𝑣﷯﷯ 𝑣+𝑥 𝑑𝑣﷮𝑑𝑥﷯= 1 − 3 𝑣﷮2﷯﷮ 𝑣﷮3﷯ −3𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 1 − 3 𝑣﷮2﷯﷮ 𝑣﷮3﷯ −3𝑣﷯ −𝑣 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 1 − 3 𝑣﷮2﷯ − 𝑣﷮4﷯ + 3 𝑣﷮2﷯﷮ 𝑣﷮3﷯ −3𝑣﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯= 1 − 𝑣﷮4﷯﷮ 𝑣﷮3﷯ − 3𝑣﷯ 𝑑𝑣﷮ 1 − 𝑣﷮4﷯﷮ 𝑣﷮3﷯ − 3𝑣﷯﷯﷯= 𝑑𝑥﷮𝑥﷯ 𝑣﷮3﷯ −3𝑣﷯𝑑𝑣﷮ 1 − 𝑣﷮4﷯﷯﷯= 𝑑𝑥﷮𝑥﷯ Integrating Both Sides ﷮﷮ 𝑣﷮3﷯ −3𝑣 ﷮1 − 𝑣﷮4﷯﷯𝑑𝑣﷯= ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮ 𝑣﷮3﷯ −3𝑣 ﷮1 − 𝑣﷮4﷯﷯𝑑𝑣﷯= log﷮|𝑥|﷯+𝐶 Let I = ﷮﷮ 𝑣﷮3﷯ − 3𝑣﷮1 − 𝑣﷮4﷯﷯﷯ 𝑑𝑣 ⇒ 𝐼 = log﷮𝑥+𝑐﷯ Solving 𝐼 𝐼 = ﷮﷮ 𝑣﷮3﷯ −3𝑣 ﷮1 − 𝑣﷮4﷯﷯𝑑𝑣﷯ = ﷮﷮ 𝑣﷮3﷯ ﷮1 − 𝑣﷮4﷯﷯−3 ﷮﷮ 𝑣﷮1 − 𝑣﷮4﷯﷯𝑑𝑣﷯﷯ 𝐼 = ﷮﷮ 𝑣﷮3﷯﷮−𝑡﷯ 𝑑𝑡﷮4 𝑣﷮3﷯﷯ −3 ﷮﷮ 𝑣﷮1 − 𝑢﷮2﷯﷯ 𝑑𝑢﷮2𝑣﷯﷯﷯ 𝐼 =− 1﷮4﷯ ﷮﷮ 𝑑𝑡﷮𝑡﷯+ 3﷮2﷯ ﷮﷮ 𝑑𝑢﷮ 𝑢﷮2﷯ − 1﷮2﷯﷯﷯﷯﷯ 𝐼 = −1﷮ 4﷯ log﷮𝑡﷯+ 3﷮2﷯ × 1﷮2(1)﷯𝑙𝑜𝑔 𝑢 − 1﷮𝑢 + 1﷯﷯ Putting t = 𝑣﷮4﷯ − 1 and u = v2 I = −1﷮4﷯ log﷮( 𝑣﷮4﷯−1) ﷯+ 3﷮4﷯ log﷮ ( 𝑣﷮2﷯ − 1)﷮( 𝑣﷮2﷯ + 1)﷯﷯ I = 1﷮4﷯ − log﷮ 𝑣﷮4﷯−1﷯+3 𝑙𝑜𝑔 ( 𝑣﷮2﷯ − 1)﷮( 𝑣﷮2﷯ + 1)﷯﷯﷯ I = 1﷮4﷯ − log﷮ 𝑣﷮4﷯−1﷯+ 𝑙𝑜𝑔 𝑣﷮2﷯ − 1﷯﷮3﷯﷮ 𝑣﷮2﷯ + 1﷯﷮3﷯﷯﷯﷯ I = 1﷮4﷯ 𝑙𝑜𝑔 𝑣﷮2﷯ − 1﷯﷮3﷯﷮ 𝑣﷮2﷯ + 1﷯﷮3﷯﷯ × 1﷮( 𝑣﷮4﷯ − 1)﷯﷯ I = 1﷮4﷯ 𝑙𝑜𝑔 𝑣﷮2﷯ − 1﷯﷮3﷯﷮ 𝑣﷮2﷯ + 1﷯﷮3﷯﷯ × 1﷮( 𝑣﷮2﷯ − 1)( 𝑣﷮2﷯ + 1)﷯﷯ I = 1﷮4﷯ 𝑙𝑜𝑔 𝑣﷮2﷯ − 1﷯﷮2﷯﷮ 𝑣﷮2﷯ + 1﷯﷮4﷯﷯ ﷯ I = 1﷮4﷯ 𝑙𝑜𝑔 𝑣﷮2﷯ − 1﷯﷮2﷯﷮ 𝑣﷮2﷯ + 1﷯﷮4﷯﷯ ﷯ I = 2﷮4﷯ log﷮ 𝑣﷮2﷯ − 1﷯﷮ 𝑣﷮2﷯ + 1﷯﷮2﷯﷯﷯ I = 1﷮2﷯ log﷮ 𝑣﷮2﷯ − 1﷯﷮ 𝑣﷮2﷯ + 1﷯﷮2﷯﷯﷯ Put v = 𝑦﷮𝑥﷯ I = 1﷮2﷯ log﷮ 𝑦﷮𝑥﷯﷯﷮2﷯ − 1﷯﷮ 𝑦﷮𝑥﷯﷯﷮2﷯ + 1﷯﷮2﷯﷯﷯ I = 1﷮2﷯ log 𝑦﷮2﷯ − 𝑥2﷮ 𝑥﷮2﷯﷯ ﷮ 𝑦﷮2﷯ + 𝑥2﷮ 𝑥﷮2﷯﷯﷯﷮2﷯﷯﷯ I = 1﷮2﷯ log 𝑥2( 𝑦﷮2﷯ − 𝑥﷮2﷯)﷮ 𝑦﷮2﷯ + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ Substituting value of I in (2) I = log |x| + c 1﷮2﷯ log 𝑥2(𝑦2 − 𝑥2)﷮(𝑦2 + 𝑥2)﷯﷯ = log |x| + c log 𝑥2(𝑦2 − 𝑥2)﷮(𝑥2 + 𝑦2)﷯﷯ = 2 log |x| + 2c log 𝑥2(𝑦2 − 𝑥2)﷮(𝑥2 + 𝑦2)﷯﷯ = 2 log |x| + log c1 log 𝑥2(𝑦2 − 𝑥2)﷮ 𝑥2 + 𝑦2﷯﷮2﷯﷯﷯ = log c1|x|2 𝑥2(𝑦2 − 𝑥2)﷮ 𝑥2 + 𝑦2﷯﷮2﷯﷯ = c1 x2 x2 (y2 − x2) = c1 x2 (x2 + y2)2 y2 − x2 = c1 (x2 + y2)2 x2 − y2 = c2 (x2 + y2)2 Hence proved

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