# Ex 9.5, 11

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.5, 11For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition :(π₯+π¦)ππ¦+(π₯βπ¦)ππ₯=0;π¦=1 When π₯=1 The differential equation can be written as (π₯+π¦)ππ¦+(π₯βπ¦)ππ₯=0 ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) Let F(x, y) = ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) Finding F(πx, πy) F(πx, πy) = (β(ππ₯ β ππ¦))/(ππ₯ + ππ¦) = (βπ(π₯ β π¦))/(π(π₯ + π¦)) = (β(π₯ β π¦))/(π₯ + π¦) = πΒ° F(x, y) β΄ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x ππ¦/ππ₯ = x ππ£/ππ₯ + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = (β(π₯ β π¦))/(π₯ + π¦) v + (π₯ ππ£)/ππ₯ = (β(π₯ β π£π₯))/(π₯ + π£π₯) v + (π₯ ππ£)/ππ₯ = (βπ₯(1 β π£))/(π₯(1 + π£)) v + (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£) (π₯ ππ£)/ππ₯ = (π£ β 1)/(1 + π£) β v (π₯ ππ£)/ππ₯ = (π£ β 1 β π£ β π£^2)/(1 + π£) (π₯ ππ£)/ππ₯ = (β(1 + π£^2 ))/(1 + π£) (1 + π£)/(1 + π£^2 ) dv = (βππ₯)/π₯ Integrating both sides β«1βγ(1 + π£)/(1 + π£^2 ) ππ£=ββ«1βππ₯/π₯γ β«1βγ1/(π£^2 + 1) ππ£+β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ γ β«1βγ(1 + π£)/(1 + π£)=ππ£=ββ«1βππ₯/π₯γ β«1βγ1/(π£^2 + 1)=ππ£+β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ γ tanβ1 v + β«1βπ£/(π£^2 + 1) ππ£=βlogβ‘γπ₯+πγ Put v2 + 1 = t 2v dv = dt v dv = ππ‘/2 Thus, our equation becomes tanβ1 v + β«1βγ1/π‘ Γ ππ‘/2 " = βlog x + c " γ tanβ1 v + 1/2 log t = βlogβ‘γπ₯+πγ Putting back value of t tanβ1 v + 1/2 log (v2 + 1) = βlogβ‘γπ₯+πγ Putting value of v = π¦/π₯ tanβ1 π¦/π₯+1/2 log (π¦^2/π₯^2 " + 1" ) = βlogβ‘γπ₯+πγ tanβ1 π¦/π₯+1/2 log (π¦^2 + π₯^2)/π₯^2 +logβ‘γπ₯=πγ tanβ1 π¦/π₯+ logβ((π₯^2 + π¦^2)/π₯^2 ) +logβ‘γπ₯=πγ tanβ1 π¦/π₯ + log β(π₯^2 + π¦^2 )/β(π₯^2 )+ log x = c tanβ1 π¦/π₯ + log β(π₯^2 + π¦^2 )/π₯+ log x = c tanβ1 π¦/π₯ + log β(π₯^2 γ+π¦γ^2 ) β log x + log x = C tanβ1 π¦/π₯ + log β(π₯^2 γ+π¦γ^2 ) = C Putting x = 1 & y = 1 tanβ1 (1/1) + log β(1^2+1^2 ) = C tanβ1 1 + log β2 = C π/4 + 1/2 log 2 = C Put value of C in (2) tanβ1 π¦/π₯ + log β(π₯^2+π¦^2 ) = π/4 + 1/2 log 2 tanβ1 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 tanβ1 π/π = π /π + log 2

Chapter 9 Class 12 Differential Equations

Serial order wise

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