Check sibling questions

Ex 9.5, 3 - Show homogeneous: (x - y) dx - (x + y) dx = 0

Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 4
Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 5
Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 6
Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.