# Example 28

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 28 (Introduction) Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) Now, ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦βπ₯γ Put F(x, y) = ππ¦/ππ₯ F(x, y) = (1 + π¦^2)/(tan^(β1)β‘π¦βπ₯) F(πx, πy) = (1 + π^2 π¦^2)/(tan^(β1)β‘ππ¦βππ₯) F(πx, πy) β πΒ° F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method ππ¦/ππ₯ = ((1 + π¦^2))/(tan^(β1) y β x) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We need to find ππ₯/ππ¦ ππ₯/ππ¦ = (tan^(β1)β‘π¦ β π₯)/(1 + π¦^2 ) ππ₯/ππ¦ = tan^(β1)β‘π¦/(1 + π¦^2 ) β π₯/(1 + π¦^2 ) ππ₯/ππ¦ + π₯/(1 + π¦^2 ) β (tan^(β1)β‘π¦ )/(1 + π¦^2 ) Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 Thus, we solve question by integrating factor method taking π π/π π Example 28 Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) IF = π^tan^(β1)β‘π¦ Solution is x (IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆγ xπ^tan^(β1)β‘π¦ = β«1β(tan^(β1)β‘γπ¦ π^tan^(β1)β‘π¦ γ )/(1 + π¦^2 ) dy + C Let I = β«1β(tan^(β1)β‘γπ¦ π^tan^(β1)β‘π¦ γ )/(1 + π¦^2 ) dy Putting tan^(β1)β‘γπ¦ γ= t 1/(1 + π¦^2 ) dy = dt Putting values of t & dt in I I = β«1βγπ‘π^π‘ ππ‘γ I = t.β«1βγπ^π‘ ππ‘ββ«1β[1β«1βγπ^π‘ ππ‘γ] ππ‘γ I = tπ^π‘ β β«1βγπ^π‘ ππ‘γ I = tπ^π‘ β π^π‘ I = π^π‘ (t β 1) Putting value of t = tan^(β1)β‘π¦ I = π^(tan^(β1)β‘π¦ ) (tan^(β1) π¦β1) Putting value of I in (1) γπ₯πγ^(tan^(β1)β‘π¦ ) = πΌ + πΆ γπ₯πγ^(tan^(β1) π¦ )= π^(tan^(β1) π¦ ) (tan^(β1)β‘π¦β1) + c Divide by tan^(β1) π¦ γππγ^(γπππγ^(βπ) π )= (γπππγ^(βπ)β‘πβπ) + cπ^(βγπππγ^(βπ) π ) Which is the required general solution

Example 1
Important

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7 Important

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13 Important

Example 14

Example 15

Example 16

Example 17 Important

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important You are here

Chapter 9 Class 12 Differential Equations

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.