# Ex 9.4, 9 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Ex 9.4, 9 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=sin^(β1)β‘π₯ ππ¦/ππ₯=sin^(β1)β‘π₯ ππ¦ = sin^(β1)β‘π₯ dx Integrating both sides β«1βγππ¦ γ= β«1βγsin^(β1)β‘γπ₯.1 ππ₯γ γ y = sinβ1 x β«1βγ1 ππ₯ ββ«1β[1/β(1 β π₯^2 ) β«1βγ1.ππ₯ γ] γ dx Integrating by parts, using formula β«1βγπ (π₯)π(π₯)ππ₯ γ= π(π₯) β«1βγπ(π₯)ππ₯ ββ«1βγ[πβ²(π₯)β«1βπ(π₯)ππ₯] ππ₯ γ γ Take f(x) = sinβ1 x and g(x) = 1 ("β΄" (πγ(sinγ^(β1) π₯))/ππ₯=1/β(1 β π₯^2 )) y = x γπ ππγ^(β1) π₯ β β«1βπ₯/β(1 β π₯^2 ) dx Let t = 1 β x2 dt = β2xdx x dx = (βππ‘)/2 Hence, our equation becomes y = x sinβ1 x β β«1β(βππ‘)/(2βπ‘) y = x sinβ1 x + β«1βππ‘/(2βπ‘) y = x sinβ1 x + 1/2 β«1βγπ‘^((β1)/2) ππ‘γ y = x sinβ1 x + 1/2 π‘^((β1)/2 + 1)/((β1)/2 + 1) + C y = x sinβ1 x + 1/2 (π‘^(1/2) )/((1/2) )+πΆ y = x sinβ1 x + βπ‘ + C Putting back value of t y = x sinβ1 x + β(πβπ^π ) + C

Ex 9.4

Ex 9.4, 1
Important

Ex 9.4, 2

Ex 9.4, 3

Ex 9.4, 4 Important

Ex 9.4, 5

Ex 9.4, 6

Ex 9.4, 7 Important

Ex 9.4, 8

Ex 9.4, 9 Important You are here

Ex 9.4, 10 Important

Ex 9.4, 11 Important

Ex 9.4, 12

Ex 9.4, 13

Ex 9.4, 14

Ex 9.4, 15 Important

Ex 9.4, 16

Ex 9.4, 17 Important

Ex 9.4, 18

Ex 9.4, 19 Important

Ex 9.4, 20 Important

Ex 9.4, 21

Ex 9.4, 22 Important

Ex 9.4, 23 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.