Ex 9.6

Chapter 9 Class 12 Differential Equations
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Transcript

Ex 9.6, 19 The integrating Factor of the differential equation (1βπ¦^2 ) ππ₯/ππ¦+π¦π₯=ππ¦ (β1<π¦<1) is (A) 1/(π¦^2β1) (B) 1/β(π¦^2β1) (C) 1/(1βπ¦^2 ) (D) 1/β(1βπ¦^2 ) (1βπ¦^2 ) ππ₯/ππ¦+π¦π₯=ππ¦ Dividing both sides by 1 β y2 ππ₯/ππ¦ + π¦π₯/(1βπ¦^2 ) = ππ¦/(1βπ¦^2 ) Differential equation is of the form ππ/ππ + P1x = Q1 where P1 = π¦/(1 β π¦^2 ) & Q1 = ππ¦/(1 β π¦^2 ) IF = π^β«1βπ1ππ¦ Finding β«1βγπ·π ππγ β«1βγπ1 ππ¦=γ β«1βγπ¦/(1βπ¦^2 ) ππ¦ γ Putting 1 β y2 = t β2y dy = dt y dy = (β1)/2 dt β΄ Our equation becomes β«1βγπ1 ππ¦= (β1)/2 γ β«1βγππ‘/π‘ γ β«1βγπ1 ππ¦= (β1)/2 γ logβ‘π‘ Putting back value of t β«1βγπ1 ππ¦= (β1)/2 γ logβ‘(1βπ¦^2) β«1βγπ1 ππ¦=γ γlogβ‘(1βπ¦^2 )γ^((β1)/2) β«1βγπ1 ππ¦= γ log 1/β(1 β π¦^2 ) Thus, IF = π^β«1βπ1ππ₯ IF = π^(πππ 1/β(1 β π¦^2 )) IF = π/β(π β π^π ) So, the correct answer is (d)