Slide32.JPG

Slide33.JPG
Slide34.JPG
Slide35.JPG
Slide36.JPG Slide37.JPG Slide38.JPG Slide39.JPG

Go Ad-free

Transcript

Ex 9.4, 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 π’…π’š/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯ = (βˆ’(π‘₯ βˆ’ 𝑦))/(π‘₯ + 𝑦) 𝒗 + (𝒙 𝒅𝒗)/𝒅𝒙 = (βˆ’(𝒙 βˆ’ 𝒗𝒙))/(𝒙 + 𝒗𝒙) 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’™)/𝒙 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〗 tanβˆ’1 v + ∫1▒𝒗/(𝒗^𝟐 + 𝟏) 𝒅𝒗=βˆ’π’π’π’ˆβ‘|𝒙|+π‘ͺ 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 π’š/𝒙 + 𝟏/πŸΓ—(π’π’π’ˆβ‘(π’š^𝟐+𝒙^𝟐 ) )=π‘ͺ 2tanβˆ’1 𝑦/π‘₯ + (log⁑(𝑦^2+π‘₯^2 ) )=2𝐢 2tanβˆ’1 𝑦/π‘₯ + (log⁑(𝑦^2+π‘₯^2 ) )=𝐾 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/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"

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo