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Ex 9.5, 11 - Find particular solution: (x + y) dy + (x-y)dx = 0

Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 5 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 6 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 7 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 8 Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations - Part 9

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Ex 9.5, 11 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : (π‘₯+𝑦)𝑑𝑦+(π‘₯βˆ’π‘¦)𝑑π‘₯=0;𝑦=1 When π‘₯=1 The differential equation can be written as (π‘₯+𝑦)𝑑𝑦+(π‘₯βˆ’π‘¦)𝑑π‘₯=0 𝑑𝑦/𝑑π‘₯ = (βˆ’(π‘₯ βˆ’ 𝑦))/(π‘₯ + 𝑦) Let F(x, y) = 𝑑𝑦/𝑑π‘₯ = (βˆ’(π‘₯ βˆ’ 𝑦))/(π‘₯ + 𝑦) Finding F(𝝀x, 𝝀y) F(πœ†x, πœ†y) = (βˆ’(πœ†π‘₯ βˆ’ πœ†π‘¦))/(πœ†π‘₯ + πœ†π‘¦) = (βˆ’πœ†(π‘₯ βˆ’ 𝑦))/(πœ†(π‘₯ + 𝑦)) = (βˆ’(π‘₯ βˆ’ 𝑦))/(π‘₯ + 𝑦) = πœ†Β° F(x, y) ∴ F(x, y) is a homogenous function of degree zero Putting y = vx. Differentiating w.r.t. x 𝑑𝑦/𝑑π‘₯ = π‘₯ 𝑑𝑣/𝑑π‘₯ + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯ = (βˆ’(π‘₯ βˆ’ 𝑦))/(π‘₯ + 𝑦) 𝑣 + (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (βˆ’(π‘₯ βˆ’ 𝑣π‘₯))/(π‘₯ + 𝑣π‘₯) 𝑣 + (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (βˆ’π‘₯(1 βˆ’ 𝑣))/(π‘₯(1 + 𝑣)) 𝑣 + (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (𝑣 βˆ’ 1)/(1 + 𝑣) (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (𝑣 βˆ’ 1)/(1 + 𝑣)βˆ’π‘£ (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (𝑣 βˆ’ 1 βˆ’ 𝑣(1 + 𝑣) )/(1 + 𝑣) (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (𝑣 βˆ’ 1 βˆ’ 𝑣 βˆ’ 𝑣^2)/(1 + 𝑣) (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (βˆ’(1 + 𝑣^2 ))/(1 + 𝑣) (1 + 𝑣)/(1 + 𝑣^2 ) 𝑑𝑣 = (βˆ’π‘‘π‘₯)/π‘₯ Integrating both sides ∫1β–’γ€–(1 + 𝑣)/(1 + 𝑣^2 ) 𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯γ€— ∫1β–’γ€–1/(𝑣^2 + 1) 𝑑𝑣+∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=βˆ’log⁑|π‘₯|+𝐢〗 ∫1β–’γ€–(1 + 𝑣)/(1 + 𝑣^2 ) 𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯γ€— ∫1β–’γ€–1/(𝑣^2 + 1) 𝑑𝑣+∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=βˆ’log⁑|π‘₯|+𝐢〗 tanβˆ’1 v + ∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=βˆ’log⁑|π‘₯|+𝐢 Putting v2 + 1 = t 2v dv = dt v dv = 𝑑𝑑/2 Thus, our equation becomes tanβˆ’1 v + ∫1β–’γ€–1/𝑑 Γ— 𝑑𝑑/2 " =" βˆ’log⁑|π‘₯|+𝐢" " γ€— tanβˆ’1 v + 1/2Γ—log⁑〖|𝑑|γ€— " ="βˆ’log⁑|π‘₯|+𝐢 Putting back value of t tanβˆ’1 v + 1/2Γ—log⁑〖|𝑣^2+1|γ€— " ="βˆ’log⁑|π‘₯|+𝐢 Putting back value of v = 𝑦/π‘₯ tanβˆ’1 𝑦/π‘₯ + 1/2Γ—log⁑|(𝑦/π‘₯)^2+1| " ="βˆ’log⁑|π‘₯|+𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—log⁑|(𝑦/π‘₯)^2+1| " "+log⁑|π‘₯|=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—log⁑|(𝑦/π‘₯)^2+1| " "+2/2 log⁑|π‘₯|=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑|(𝑦/π‘₯)^2+1| " " +2 log⁑|π‘₯| )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑|𝑦^2/π‘₯^2 +1| " " +log⁑|π‘₯^2 | )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑((𝑦^2 + π‘₯^2)/π‘₯^2 )" " +log⁑〖π‘₯^2 γ€— )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑((𝑦^2 + π‘₯^2)/π‘₯^2 Γ—π‘₯^2 ) )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑(𝑦^2+π‘₯^2 ) )=𝐢 2tanβˆ’1 𝑦/π‘₯ + (log⁑(𝑦^2+π‘₯^2 ) )=2𝐢 2tanβˆ’1 𝑦/π‘₯ + (log⁑(𝑦^2+π‘₯^2 ) )=𝐾 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑|(𝑦/π‘₯)^2+1| " " +2 log⁑|π‘₯| )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑|𝑦^2/π‘₯^2 +1| " " +log⁑|π‘₯^2 | )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑((𝑦^2 + π‘₯^2)/π‘₯^2 )" " +log⁑〖π‘₯^2 γ€— )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑((𝑦^2 + π‘₯^2)/π‘₯^2 Γ—π‘₯^2 ) )=𝐢 tanβˆ’1 𝑦/π‘₯ + 1/2Γ—(log⁑(𝑦^2+π‘₯^2 ) )=𝐢 2tanβˆ’1 𝑦/π‘₯ + log⁑(𝑦^2+π‘₯^2 )=2𝐢 2tanβˆ’1 𝑦/π‘₯ + log⁑(𝑦^2+π‘₯^2 )=𝐾 …(2) Now, Putting x = 1 & y = 1 in equation 2tanβˆ’1 𝑦/π‘₯ + log⁑(𝑦^2+π‘₯^2 )=𝐾 2tanβˆ’1 (1/1) + log (1^2+1^2 )= K 2tanβˆ’1 1 + log 2 = K 2 Γ— πœ‹/4 + log 2 = K πœ‹/2 + log 2 = K K = πœ‹/2 + log 2 Put value of K in (2) 2tanβˆ’1 𝑦/π‘₯ + log⁑(𝑦^2+π‘₯^2 )=𝐾 2tanβˆ’1 𝑦/π‘₯ + log⁑(𝑦^2+π‘₯^2 )=πœ‹/2 " + log 2" π’π’π’ˆβ‘(π’š^𝟐+𝒙^𝟐 )+ 2tanβˆ’1 π’š/𝒙 =𝝅/𝟐 " + log 2"

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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.