# Example 28 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Example 28 (Introduction) Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) ππ₯/ππ¦+π₯/(1 + π¦^2 )=tan^(β1)β‘π¦/(1 + π¦^2 ) ππ₯/ππ¦=tan^(β1)β‘π¦/(1 + π¦^2 ) β π₯/(1 + π¦^2 ) ππ₯/ππ¦=(tan^(β1)β‘π¦ β π₯)/(1 + π¦^2 ) ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦ β π₯γ The variables cannot be separated. So variable separable method is not possible Now, ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦ β π₯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₯)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 ) ππ₯/ππ¦+π₯/(1 + π¦^2 )=tan^(β1)β‘π¦/(1 + π¦^2 ) Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = 1/(1 + π¦^2 ) & Q1 = (tan^(β1)β‘π¦ )/(1 + π¦^2 ) Now, IF = π^β«1βγπ_1 ππ¦γ IF = π^β«1βγ1/(1 + π¦^2 ) ππ¦γ IF = π^tan^(β1)β‘π¦ Solution is x (IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆγ xπ^tan^(β1)β‘π¦ = β«1βγtan^(β1)β‘π¦/(1 + π¦^2 )Γπ^tan^(β1)β‘π¦ ππ¦γ + C Let I = β«1βγtan^(β1)β‘π¦/(1 + π¦^2 )Γπ^tan^(β1)β‘π¦ ππ¦γ 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 Integrating by parts with β«1βγπ(π‘) π(π‘) ππ‘=π(π‘) β«1βγπ(π‘) ππ‘ ββ«1βγ[π^β² (π‘) β«1βγπ(π‘) ππ‘] ππ‘γγγγ Take f (t) = t & g (t) = π^"t" Divide by π^(γπ‘ππγ^(β1) π¦ ) γπ₯πγ^(tan^(β1) π¦ )/π^tan^(β1)β‘γπ¦ γ = (π^(tan^(β1) π¦) (tan^(β1)β‘π¦ β 1))/π^tan^(β1)β‘γπ¦ γ + π/π^tan^(β1)β‘γπ¦ γ π = (γπππγ^(βπ)β‘πβπ) + cπ^(βγπππγ^(βπ) π ) Which is the required general solution

Examples

Example 1
Important

Example 2

Example 3

Example 4 Deleted for CBSE Board 2021 Exams only

Example 5 Deleted for CBSE Board 2021 Exams only

Example 6 Important Deleted for CBSE Board 2021 Exams only

Example 7 Deleted for CBSE Board 2021 Exams only

Example 8 Deleted for CBSE Board 2021 Exams only

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18

Example 19

Example 20

Example 21 Deleted for CBSE Board 2021 Exams only

Example 22 Important

Example 23 Important

Example 24

Example 25 Deleted for CBSE Board 2021 Exams only

Example 26

Example 27 Important

Example 28 Important Deleted for CBSE Board 2021 Exams only 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 10 years. He provides courses for Maths and Science at Teachoo.