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

Transcript

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's photo - Teacher, Computer Engineer, Marketer
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.