Example 22 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving Linear differential equations - Equation given
Ex 9.5, 19 (MCQ) Important
Misc 14 (MCQ) Important
Ex 9.5, 2
Ex 9.5, 10
Ex 9.5, 3 Important
Ex 9.5, 4
Misc 15 (MCQ)
Ex 9.5, 13
Ex 9.5, 8 Important
Misc 10 Important
Misc 11
Ex 9.5, 14 Important
Ex 9.5, 6
Ex 9.5, 5 Important
Ex 9.5, 9
Ex 9.5, 7 Important
Ex 9.5, 15
Example 14
Ex 9.5, 1 Important
Ex 9.5, 12 Important
Ex 9.5, 11
Example 16
Example 17 Important
Example 22 Important You are here
Solving Linear differential equations - Equation given
Last updated at April 16, 2024 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