





Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Examples
Example 1 (ii) Important
Example 1 (iii) Important
Example 2
Example 3 Important
Example 4
Example 5
Example 6
Example 7 Important
Example 8
Example 9 Important
Example 10 Important
Example 11
Example 12 Important
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19
Example 20
Example 21 Important
Example 22 Important You are here
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Last updated at Aug. 14, 2023 by Teachoo
Example 22 (Introduction) Solve the differential equation ๐๐ฅ/๐๐ฆ+๐ฅ/(1+๐ฆ^2 )=tan^(โ1)โก๐ฆ/(1+๐ฆ^2 ) ๐๐ฅ/๐๐ฆ+๐ฅ/(1 + ๐ฆ^2 )=tan^(โ1)โก๐ฆ/(1 + ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ=tan^(โ1)โก๐ฆ/(1 + ๐ฆ^2 ) โ ๐ฅ/(1 + ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ=(tan^(โ1)โก๐ฆ โ ๐ฅ)/(1 + ๐ฆ^2 ) ๐ ๐/๐ ๐ = ((๐ + ๐^๐))/ใ๐๐๐ใ^(โ๐)โกใ๐ โ ๐ใ The variables cannot be separated. So variable separable method is not possible Now, ๐๐ฆ/๐๐ฅ = ((1 + ๐ฆ^2))/tan^(โ1)โกใ๐ฆโ๐ฅใ Put F(x, y) = ๐ ๐/๐ ๐ F(x, y) = (1 + ๐ฆ^2)/(tan^(โ1)โก๐ฆโ๐ฅ) F(๐x, ๐y) = (1 + ๐^2 ๐ฆ^2)/(tan^(โ1)โก๐๐ฆโ๐๐ฅ) F(๐x, ๐y) โ ๐ยฐ F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method ๐๐ฆ/๐๐ฅ = ((1 + ๐ฆ^2))/(tan^(โ1) y โ x) This is not of the form ๐ ๐/๐ ๐+๐ท๐=๐ธ โด We need to find ๐ ๐/๐ ๐ ๐๐ฅ/๐๐ฆ = (tan^(โ1)โก๐ฆ โ ๐ฅ)/(1 + ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ = tan^(โ1)โก๐ฆ/(1 + ๐ฆ^2 ) โ ๐ฅ/(1 + ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ + ๐ฅ/(1 + ๐ฆ^2 ) โ (tan^(โ1)โก๐ฆ )/(1 + ๐ฆ^2 ) Differential equation is of the form ๐ ๐/๐ ๐ + P1 x = Q1 Thus, we solve question by integrating factor method taking ๐ ๐/๐ ๐ Example 22 Solve the differential equation ๐๐ฅ/๐๐ฆ+๐ฅ/(1+๐ฆ^2 )=tan^(โ1)โก๐ฆ/(1+๐ฆ^2 ) ๐๐ฅ/๐๐ฆ+๐ฅ/(1 + ๐ฆ^2 )=tan^(โ1)โก๐ฆ/(1 + ๐ฆ^2 ) Differential equation is of the form ๐ ๐/๐ ๐ + P1 x = Q1 where P1 = 1/(1 + ๐ฆ^2 ) & Q1 = (tan^(โ1)โก๐ฆ )/(1 + ๐ฆ^2 ) Now, IF = ๐^โซ1โใ๐_1 ๐๐ฆใ IF = ๐^โซ1โใ1/(1 + ๐ฆ^2 ) ๐๐ฆใ IF = ๐^ใ๐๐๐ใ^(โ๐)โก๐ Solution is x (IF) = โซ1โใ(๐ร๐ผ๐น)๐๐ฆ+๐ถใ x๐^ใ๐๐๐ใ^(โ๐)โก๐ = โซ1โใใ๐๐๐ใ^(โ๐)โก๐/(๐ + ๐^๐ )ร๐^ใ๐๐๐ใ^(โ๐)โก๐ ๐ ๐ใ + C Let I = โซ1โใtan^(โ1)โก๐ฆ/(1 + ๐ฆ^2 )ร๐^tan^(โ1)โก๐ฆ ๐๐ฆใ Let ใ๐๐๐ใ^(โ๐)โกใ๐ ใ= t 1/(1 + ๐ฆ^2 ) dy = dt Putting values of t & dt in I I = โซ1โใ๐๐^๐ ๐ ๐ใ Integrating by parts with โซ1โใ๐(๐ก) ๐(๐ก) ๐๐ก=๐(๐ก) โซ1โใ๐(๐ก) ๐๐ก โโซ1โใ[๐^โฒ (๐ก) โซ1โใ๐(๐ก) ๐๐ก] ๐๐กใใใใ Take f (t) = t & g (t) = ๐^"t" I = t.โซ1โใ๐^๐ก ๐๐กโโซ1โ[1โซ1โใ๐^๐ก ๐๐กใ] ๐๐กใ I = t๐^๐ก โ โซ1โใ๐^๐ก ๐๐กใ I = t๐^๐ โ ๐^๐ I = ๐^๐ก (t โ 1) Putting value of t = tan^(โ1)โก๐ฆ I = ๐^(ใ๐ญ๐๐งใ^(โ๐)โก๐ ) (ใ๐๐๐ใ^(โ๐) ๐โ๐) Putting value of I in (1) ใ๐ฅ๐ใ^(tan^(โ1)โก๐ฆ ) = ๐ผ + ๐ถ ใ๐ฅ๐ใ^(tan^(โ1) ๐ฆ )= ๐^(tan^(โ1) ๐ฆ ) (tan^(โ1)โก๐ฆโ1) + c Divide by ๐^(ใ๐๐๐ใ^(โ๐) ๐ ) ใ๐ฅ๐ใ^(tan^(โ1) ๐ฆ )/๐^tan^(โ1)โกใ๐ฆ ใ = (๐^(tan^(โ1) ๐ฆ) (tan^(โ1)โก๐ฆ โ 1))/๐^tan^(โ1)โกใ๐ฆ ใ + ๐/๐^tan^(โ1)โกใ๐ฆ ใ ๐ = (ใ๐๐๐ใ^(โ๐)โก๐โ๐) + c๐^(โใ๐๐๐ใ^(โ๐) ๐ ) Which is the required general solution