Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### 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₯+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. Differentiating w.r.t. x ππ/ππ = π 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¦) π + (π ππ)/ππ = (β(π β ππ))/(π + ππ) π£ + (π₯ ππ£)/ππ₯ = (βπ₯(1 β π£))/(π₯(1 + π£)) π£ + (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£) (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£)βπ£ (π₯ ππ£)/ππ₯ = (π£ β 1 β π£(1 + π£) )/(1 + π£) (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£) (π + π)/(π + π^π ) ππ = (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‘|π₯|+πΆγ tanβ1 v + β«1βπ/(π^π + π) ππ=βπππβ‘|π|+πͺ Putting 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logβ‘|π₯|+πΆ" " γ tanβ1 v + 1/2Γlogβ‘γ|π‘|γ " ="βlogβ‘|π₯|+πΆ Putting back value of t tanβ1 v + 1/2Γlogβ‘γ|π£^2+1|γ " ="βlogβ‘|π₯|+πΆ Putting back value of v = π¦/π₯ tanβ1 π¦/π₯ + 1/2Γlogβ‘|(π¦/π₯)^2+1| " ="βlogβ‘|π₯|+πΆ tanβ1 π¦/π₯ + 1/2Γlogβ‘|(π¦/π₯)^2+1| " "+logβ‘|π₯|=πΆ tanβ1 π¦/π₯ + 1/2Γlogβ‘|(π¦/π₯)^2+1| " "+2/2 logβ‘|π₯|=πΆ tanβ1 π¦/π₯ + 1/2Γ(logβ‘|(π¦/π₯)^2+1| " " +2 logβ‘|π₯| )=πΆ tanβ1 π¦/π₯ + 1/2Γ(logβ‘|π¦^2/π₯^2 +1| " " +logβ‘|π₯^2 | )=πΆ tanβ1 π¦/π₯ + 1/2Γ(logβ‘((π¦^2 + π₯^2)/π₯^2 )" " +logβ‘γπ₯^2 γ )=πΆ tanβ1 π¦/π₯ + 1/2Γ(logβ‘((π¦^2 + π₯^2)/π₯^2 Γπ₯^2 ) )=πΆ tanβ1 π/π + π/πΓ(πππβ‘(π^π+π^π ) )=πͺ 2tanβ1 π¦/π₯ + (logβ‘(π¦^2+π₯^2 ) )=2πΆ 2tanβ1 π¦/π₯ + (logβ‘(π¦^2+π₯^2 ) )=πΎ Now, Putting x = 1 & y = 1 in equation 2tanβ1 π¦/π₯ + logβ‘(π¦^2+π₯^2 )=πΎ 2tanβ1 (1/1) + log (1^2+1^2 )= K 2tanβ1 1 + log 2 = K 2 Γ π/4 + log 2 = K π/2 + log 2 = K K = π/π + log 2 Put value of K in (2) 2tanβ1 π¦/π₯ + logβ‘(π¦^2+π₯^2 )=πΎ 2tanβ1 π¦/π₯ + logβ‘(π¦^2+π₯^2 )=π/2 " + log 2" πππβ‘(π^π+π^π )+ 2tanβ1 π/π =π/π " + log 2"