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

Transcript

Ex 9.5, 6 Show that the given differential equation is homogeneous and solve each of them. ๐‘ฅ ๐‘‘๐‘ฆโˆ’๐‘ฆ ๐‘‘๐‘ฅ=โˆš(๐‘ฅ^2+๐‘ฆ^2 ) ๐‘‘๐‘ฅ Step 1: Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ x dy โˆ’ y dx = โˆš(๐‘ฅ^2+๐‘ฆ^2 ) dx x dy = โˆš(๐‘ฅ^2+๐‘ฆ^2 ) dx + ydx x dy = (โˆš(๐‘ฅ^2+๐‘ฆ^2 )+๐‘ฆ) dx ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆš(๐‘ฅ^2 + ๐‘ฆ^2 ) + ๐‘ฆ)/๐‘ฅ Step 2: Put ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = F(x, y) and find F(๐œ†x, ๐œ†y) F(x, y) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆš(๐‘ฅ^(2 )+ ๐‘ฆ^2 ) + ๐‘ฆ)/๐‘ฅ F(๐œ† x, ๐œ†y) = (โˆš(ใ€–(๐œ†๐‘ฅ)ใ€—^2 + (๐œ†^2 ๐‘ฆ^2 ) )+ ๐œ†๐‘ฆ)/๐œ†๐‘ฅ = (โˆš(๐œ†^2 ๐‘ฅ^2 + ๐œ†^2 ๐‘ฆ^2 ) + ๐œ†๐‘ฆ)/๐œ†๐‘ฅ = (โˆš(๐œ†^2 (๐‘ฅ^2 + ๐‘ฆ^2)) + ๐œ†๐‘ฆ)/๐œ†๐‘ฅ= (๐œ†โˆš(๐‘ฅ^2 + ๐‘ฆ^2 ) + ๐œ†๐‘ฆ)/๐œ†๐‘ฅ = (โˆš(๐‘ฅ^2 + ๐‘ฆ^2 ) + ๐‘ฆ)/๐‘ฅ = F(x, y) Hence, F(๐œ†x, ๐œ†y) = F(x, y) = ๐œ†ยฐ F(x, y) Hence, F(x, y) is a homogenous Function of with degree 0 So, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ is a homogenous differential equation. Step 3 - Solving ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ by putting y = vx Putting y = vx. Differentiating w.r.t.x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘ฃ Putting value of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and y = vx in (1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆš(๐‘ฅ^2 + ๐‘ฆ^2 )+ ๐‘ฆ)/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ=(โˆš(๐‘ฅ^2 + (๐‘ฃ๐‘ฅ)^2 ) + (๐‘ฃ๐‘ฅ))/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ=(โˆš(๐‘ฅ^2 + ๐‘ฅ^2 ๐‘ฃ^2 ) + ๐‘ฃ๐‘ฅ)/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =(โˆš(๐‘ฅ^2 (1 + ๐‘ฃ^2)) + ๐‘ฃ๐‘ฅ)/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =(๐‘ฅโˆš(1 + ๐‘ฃ^2 ) + ๐‘ฃ๐‘ฅ)/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =(๐‘ฅ(โˆš(1 + ๐‘ฃ^2 ) + ๐‘ฃ))/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ= โˆš(1+๐‘ฃ^2 )+๐‘ฃ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ= โˆš(1+๐‘ฃ^2 )+๐‘ฃ โˆ’ ๐‘ฃ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ= โˆš(1+๐‘ฃ^2 ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ= โˆš(1 + ๐‘ฃ^2 )/๐‘ฅ ๐‘‘๐‘ฃ/โˆš(1 + ๐‘ฃ^2 )= ๐‘‘๐‘ฅ/๐‘ฅ Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฃ/โˆš(1 + ๐‘ฃ^2 ) = โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆซ1โ–’๐‘‘๐‘ฃ/โˆš(1 + ๐‘ฃ^2 ) = log |๐‘ฅ|+๐‘ We know that โˆซ1โ–’๐‘‘๐‘ฃ/โˆš(๐‘Ž^2 + ๐‘ฅ^2 ) =๐‘™๐‘œ๐‘”|๐‘ฅ+โˆš(๐‘ฅ^2+๐‘Ž^2 )|+๐‘ Putting a = 1, x = v log |๐‘ฃ+โˆš(๐‘ฃ^2+1)| =๐‘™๐‘œ๐‘”|๐‘ฅ|+๐‘ log |๐‘ฃ+โˆš(๐‘ฃ^2+1)| =๐‘™๐‘œ๐‘”|๐‘๐‘ฅ| (As log ๐‘Ž + log b = log ab) v + โˆš(๐‘ฃ^2+1) = cx Putting v = ๐‘ฆ/๐‘ฅ ๐‘ฆ/๐‘ฅ+โˆš((๐‘ฆ/๐‘ฅ)^2+1)=๐‘๐‘ฅ ๐‘ฆ/๐‘ฅ+โˆš(๐‘ฆ^2/๐‘ฅ^2 +1)=๐‘๐‘ฅ ๐‘ฆ/๐‘ฅ+โˆš((๐‘ฆ^2 + ๐‘ฅ^2)/๐‘ฅ^2 )=๐‘๐‘ฅ ๐‘ฆ/๐‘ฅ+โˆš(๐‘ฆ^2 + ๐‘ฅ^2 )/๐‘ฅ=๐‘๐‘ฅ ๐’š+โˆš(๐’š^๐Ÿ ใ€–+ ๐’™ใ€—^๐Ÿ ) =๐’„๐’™^๐Ÿ โˆด General solution is ๐’š+โˆš(๐’š^๐Ÿ ใ€–+ ๐’™ใ€—^๐Ÿ ) =๐’„๐’™^๐Ÿ

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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.