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₯+π¦)ππ¦+(π₯βπ¦)ππ₯=0;π¦=1 When π₯=1 The differential equation can be written as (π₯+π¦)ππ¦+(π₯βπ¦)ππ₯=0 ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) Let F(x, y) = ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) Finding F(πx, πy) F(πx, πy) = (β(ππ₯ β ππ¦))/(ππ₯ + ππ¦) = (βπ(π₯ β π¦))/(π(π₯ + π¦)) = (β(π₯ β π¦))/(π₯ + π¦) = πΒ° F(x, y) β΄ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x ππ¦/ππ₯ = x ππ£/ππ₯ + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) v + (π₯ ππ£)/ππ₯ = (β(π₯ β π£π₯))/(π₯ + π£π₯) v + (π₯ ππ£)/ππ₯ = (βπ₯(1 β π£))/(π₯(1 + π£)) v + (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£) (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£) β v (π₯ ππ£)/ππ₯ = (π£ β 1 β π£ β π£^2)/(1 + π£) (π₯ ππ£)/ππ₯ = (β(1 + π£^2 ))/(1 + π£) (1 + π£)/(1 + π£^2 ) dv = (βππ₯)/π₯ Integrating both sides β«1βγ(1 + π£)/(1 + π£^2 ) ππ£=ββ«1βππ₯/π₯γ β«1βγ1/(π£^2 + 1) ππ£+β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ γ β«1βγ(1 + π£)/(1 + π£)=ππ£=ββ«1βππ₯/π₯γ β«1βγ1/(π£^2 + 1)=ππ£+β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ γ tanβ1 v + β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ Put v2 + 1 = t 2v dv = dt v dv = ππ‘/2 Thus, our equation becomes tanβ1 v + β«1βγ1/π‘ Γ ππ‘/2 " = βlog x + c " γ tanβ1 v + 1/2 log t = βlogβ‘γπ₯+πγ Putting back value of t tanβ1 v + 1/2 log (v2 + 1) = βlogβ‘γπ₯+πγ Putting value of v = π¦/π₯ tanβ1 π¦/π₯+1/2 log (π¦^2/π₯^2 " + 1" ) = βlogβ‘γπ₯+πγ tanβ1 π¦/π₯+1/2 log (π¦^2 + π₯^2)/π₯^2 +logβ‘γπ₯=πγ tanβ1 π¦/π₯+ logβ((π₯^2 + π¦^2)/π₯^2 ) +logβ‘γπ₯=πγ tanβ1 π¦/π₯ + log β(π₯^2 + π¦^2 )/β(π₯^2 )+ log x = c tanβ1 π¦/π₯ + log β(π₯^2 + π¦^2 )/π₯+ log x = c tanβ1 π¦/π₯ + log β(π₯^2 γ+π¦γ^2 ) β log x + log x = C tanβ1 π¦/π₯ + log β(π₯^2 γ+π¦γ^2 ) = C Putting x = 1 & y = 1 tanβ1 (1/1) + log β(1^2+1^2 ) = C tanβ1 1 + log β2 = C π/4 + 1/2 log 2 = C Put value of C in (2) tanβ1 π¦/π₯ + log β(π₯^2+π¦^2 ) = π/4 + 1/2 log 2 tanβ1 π¦/π₯ + 1/2 log (π₯^2+π¦^2) = π/4 + 1/2 log 2 Multiplying both sides by 2 log (π^π+π^π) + 2 tanβ1 π/π = π/π + log 2