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

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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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(πœ†x, πœ†y) 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 Diff w.r.t. x 𝑑𝑦/𝑑π‘₯ = x 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’ 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‘₯+𝑐〗 γ€— ∫1β–’γ€–(1 + 𝑣)/(1 + 𝑣)=𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯γ€— ∫1β–’γ€–1/(𝑣^2 + 1)=𝑑𝑣+∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=βˆ’log⁑〖π‘₯+𝑐〗 γ€— tanβˆ’1 v + ∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=βˆ’log⁑〖π‘₯+𝑐〗 Put 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 x + c " γ€— tanβˆ’1 v + 1/2 log t = βˆ’log⁑〖π‘₯+𝑐〗 Putting back value of t tanβˆ’1 v + 1/2 log (v2 + 1) = βˆ’log⁑〖π‘₯+𝑐〗 Putting value of v = 𝑦/π‘₯ tanβˆ’1 𝑦/π‘₯+1/2 log (𝑦^2/π‘₯^2 " + 1" ) = βˆ’log⁑〖π‘₯+𝑐〗 tanβˆ’1 𝑦/π‘₯+1/2 log (𝑦^2 + π‘₯^2)/π‘₯^2 +log⁑〖π‘₯=𝑐〗 tanβˆ’1 𝑦/π‘₯+ log√((π‘₯^2 + 𝑦^2)/π‘₯^2 ) +log⁑〖π‘₯=𝑐〗 tanβˆ’1 𝑦/π‘₯ + log √(π‘₯^2 + 𝑦^2 )/√(π‘₯^2 )+ log x = c tanβˆ’1 𝑦/π‘₯ + log √(π‘₯^2 + 𝑦^2 )/π‘₯+ log x = c tanβˆ’1 𝑦/π‘₯ + log √(π‘₯^2 γ€–+𝑦〗^2 ) βˆ’ log x + log x = C tanβˆ’1 𝑦/π‘₯ + log √(π‘₯^2 γ€–+𝑦〗^2 ) = C Putting x = 1 & y = 1 tanβˆ’1 (1/1) + log √(1^2+1^2 ) = C tanβˆ’1 1 + log √2 = C πœ‹/4 + 1/2 log 2 = C Put value of C in (2) tanβˆ’1 𝑦/π‘₯ + log √(π‘₯^2+𝑦^2 ) = πœ‹/4 + 1/2 log 2 tanβˆ’1 𝑦/π‘₯ + 1/2 log (π‘₯^2+𝑦^2) = πœ‹/4 + 1/2 log 2 Multiplying both sides by 2 log (𝒙^𝟐+π’š^𝟐) + 2 tanβˆ’1 π’š/𝒙 = 𝝅/𝟐 + log 2

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