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

Transcript

Ex 9.5, 3 In each of the Exercise 1 to 10, show that the given differential equation is homogeneous and solve each of them. (π‘₯βˆ’π‘¦)π‘‘π‘¦βˆ’(π‘₯+𝑦)𝑑π‘₯=0 Step 1: Find 𝑑𝑦/𝑑π‘₯ (x βˆ’ y) dy βˆ’ (x + y) dx = 0 (x βˆ’ y) dy = (x + y) dx 𝑑𝑦/𝑑π‘₯ = (π‘₯ + 𝑦)/(π‘₯ βˆ’ 𝑦) Step 2: Put 𝑑𝑦/𝑑π‘₯ = F(x, y) and find out F(πœ†x, πœ†y) F(x, y) = (π‘₯ + 𝑦)/(π‘₯ βˆ’ 𝑦) …(1) F(πœ†x, πœ†y) = (πœ†π‘₯ + πœ†π‘¦)/(πœ†π‘₯βˆ’πœ†π‘¦) = (πœ†(π‘₯ + 𝑦))/(πœ† (π‘₯ βˆ’ 𝑦)) = (π‘₯ + 𝑦)/(π‘₯ βˆ’ 𝑦) = F(x, y) ∴ F(πœ†x, πœ†y) = πœ†0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦/𝑑π‘₯ is a homogenous differential equation. Step 3: Solving 𝑑𝑦/𝑑π‘₯ by putting y = vx Putting y = vx. Differentiating w.r.t. x 𝑑𝑦/𝑑π‘₯ = x 𝑑𝑣/𝑑π‘₯+𝑣𝑑π‘₯/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯ 𝑑𝑣/𝑑π‘₯ + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯ = (π‘₯ + 𝑦)/(π‘₯ βˆ’ 𝑦) x 𝑑𝑣/𝑑π‘₯ + v = (π‘₯ + π‘₯𝑣)/(π‘₯ βˆ’ π‘₯𝑣) x 𝑑𝑣/𝑑π‘₯ + v = (π‘₯ (1 + 𝑣))/(π‘₯(1 βˆ’ 𝑣)) x 𝑑𝑣/𝑑π‘₯ + v = ((1 + 𝑣))/((1 βˆ’ 𝑣)) x 𝑑𝑣/𝑑π‘₯ = ((1 + 𝑣))/((1 βˆ’ 𝑣)) βˆ’ v x 𝑑𝑣/𝑑π‘₯ = (1 + 𝑣 βˆ’ 𝑣(1 βˆ’ 𝑣))/(1 βˆ’ 𝑣) x 𝑑𝑣/𝑑π‘₯ = (1 + 𝑣 βˆ’ 𝑣 + 𝑣^2)/(1 βˆ’ 𝑣) x 𝑑𝑣/𝑑π‘₯ = (1 + 𝑣^2)/(1 βˆ’ 𝑣) (1 βˆ’ 𝑣)𝑑𝑣/(1 + 𝑣^2 ) = 𝑑π‘₯/π‘₯ Integrating both sides ∫1β–’((1 βˆ’ 𝑣)/(1 + 𝑣^2 )) 𝑑𝑣=∫1▒𝑑π‘₯/π‘₯ ∫1▒𝑑𝑣/(1 + 𝑣^2 )βˆ’βˆ«1β–’(𝑣 𝑑𝑣)/(1 + 𝑣^2 )=∫1▒𝑑π‘₯/π‘₯ tanβˆ’1 v βˆ’ ∫1▒𝑣/(1 + 𝑣^2 ) = log|π‘₯|+𝑐 Let I = ∫1▒𝑣/(1 + 𝑣^2 ) dv …(2) Putting t = 1 + 𝑣^2 Diff w.r.t. v 𝑑/𝑑𝑣(1 + v2) = 𝑑𝑑/𝑑𝑣 2v = 𝑑𝑑/𝑑𝑣 dv = 𝑑𝑑/2𝑣 Therefore I = ∫1▒𝑣/(1 + 𝑣^2 ) dv = ∫1▒𝑑𝑑/2𝑑 = 1/2 π‘™π‘œπ‘”|𝑑|+𝑐 Putting back t = 1 + v2 = 1/2 π‘™π‘œπ‘”|1+𝑣^2 | + c (As ∫1β–’1/(1 + π‘₯^2 ) dx = tanβˆ’1 x) …(2) Putting value of I in (2) tanβˆ’1 v βˆ’ ∫1▒𝒗/(𝟏 + 𝒗^𝟐 ) = log|π‘₯|+𝑐 tanβˆ’1 v "βˆ’ " 𝟏/𝟐 log |𝟏+π’—πŸ| = log |π‘₯| + c tanβˆ’1 v "= " 1/2 log |1+𝑣2| + log |π‘₯| + c tanβˆ’1 v "= " 1/2 log |1+𝑣2| + 2/2 log |π‘₯| + c tanβˆ’1 v "= " 1/2 ["log " |1+𝑣2|" + " 2" log " |π‘₯|] + c tanβˆ’1 v "= " 1/2 "log" [" " |1+𝑣2|.|π‘₯|^2 ] + c Putting v = 𝑦/π‘₯ tanβˆ’1 𝑦/π‘₯ "= " 1/2 "log" [" " (1+(𝑦/π‘₯)^2 )Γ—π‘₯^2 ]+c tanβˆ’1 𝑦/π‘₯ "= " 1/2 "log" [(π‘₯^2 + 𝑦^2)/π‘₯^2 Γ—π‘₯^2 ]+c tanβˆ’1 𝑦/π‘₯ "= " 1/2 "log" [π‘₯^2+𝑦^2 ]+c tanβˆ’1 π’š/𝒙 "= " 𝟏/𝟐 "log" [𝒙^𝟐+π’š^𝟐 ]+𝐜 is the required solution

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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.