# 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 equationππ₯/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 28Solve the differential equationππ₯/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

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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.