Misc 3 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving homogeneous differential equation
Ex 9.4, 16 (MCQ)
Ex 9.4, 2
Ex 9.4, 15
Ex 9.4, 4 Important
Example 11
Ex 9.4, 14 Important
Ex 9.4, 13
Ex 9.4, 8
Ex 9.4, 7 Important
Example 21 Important
Ex 9.4, 6 Important
Ex 9.4, 5
Example 13 Important
Ex 9.4, 9
Ex 9.4, 12 Important
Ex 9.4, 1 Important
Ex 9.4, 3
Ex 9.4, 11
Example 10 Important
Misc 3 Important You are here
Example 12 Important
Ex 9.4, 10 Important
Misc 8 Important
Misc 9
Solving homogeneous differential equation
Last updated at April 16, 2024 by Teachoo
Misc 3 Prove that ๐ฅ^2โ๐ฆ^2=๐(๐ฅ^2+๐ฆ^2 )^2 is the general solution of differential equation (๐ฅ^3โ3๐ฅ๐ฆ^2 )๐๐ฅ=(๐ฆ^3โ3๐ฅ^2 ๐ฆ)๐๐ฆ, where ๐ is a parameter .Given differential equation (๐ฅ^3โ3๐ฅ๐ฆ^2 )๐๐ฅ=(๐ฆ^3โ3๐ฅ^2 ๐ฆ)๐๐ฆ (๐ฅ^3 โ 3๐ฅ๐ฆ^2)/(๐ฆ^3 โ 3๐ฅ^2 ๐ฆ)=๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=(๐ฅ^3 โ 3๐ฅ๐ฆ^2)/(๐ฆ^(3 )โ 3๐ฅ^2 ๐ฆ) ๐๐ฆ/๐๐ฅ=(๐ฅ^3 (1 โ (3๐ฅ๐ฆ^2)/๐ฅ^3 ))/(๐ฆ^(3 ) (1 โ(3๐ฅ^2 ๐ฆ)/๐ฆ^3 ) ) ๐๐ฆ/๐๐ฅ=(๐ฅ^3 (1 โ (3๐ฆ^2)/๐ฅ^2 ))/(๐ฆ^(3 ) (1 โ(3๐ฅ^2)/๐ฆ^2 ) ) ๐ ๐/๐ ๐=(๐/๐)^๐ร((๐ โ ๐(๐/๐)^๐ ))/((๐ โ ๐(๐/๐)^๐ ) ) Putting y = vx. Differentiating w.r.t. x ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ + ๐ฃ Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ =(1/๐ฃ)^3ร((1 โ 3๐ฃ^2 ))/((1 โ 3(1/๐ฃ)^2 ) ) ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ =1/๐ฃ^3 ร((1 โ 3๐ฃ^2 ))/(((๐ฃ^2 โ 3)/๐ฃ^2 ) ) ๐ ๐ ๐/๐ ๐+๐ =๐/๐ร((๐ โ ๐๐^๐ ))/((๐^๐ โ ๐) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร((1 โ 3๐ฃ^2 ))/((๐ฃ^2 โ 3) )โ๐ฃ ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร((1 โ 3๐ฃ^2 ) โ ๐ฃ ร ๐ฃ (๐ฃ^2 โ 3))/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร(1 โ 3๐ฃ^2 โ ๐ฃ^4 + 3๐ฃ^2)/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร(1 โ ๐ฃ^4)/((๐ฃ^2 โ 3) ) ๐ ๐ ๐/๐ ๐=(๐ โ ๐^๐)/((๐^๐ โ ๐๐) ) (๐^๐ โ๐๐)๐ ๐/((๐ โ๐^๐ ) )=๐ ๐/๐ Integrating Both Sides โซ1โใ(๐ฃ^3 โ3๐ฃ )/(1 โ ๐ฃ^4 ) ๐๐ฃใ=โซ1โ๐๐ฅ/๐ฅ โซ1โใ(๐^๐ โ๐๐ )/(๐ โ ๐^๐ ) ๐ ๐ใ=๐ฅ๐จ๐ โกใ|๐|ใ+๐ช Let I = โซ1โ(๐^๐ โ ๐๐)/(๐ โ ๐^๐ ) ๐ ๐ Therefore, ๐ผ =logโกใ|๐ฅ|+๐ใ Solving ๐ฐ ๐ผ =โซ1โใ(๐ฃ^3 โ3๐ฃ )/(1 โ ๐ฃ^4 ) ๐๐ฃใ =โซ1โใ(๐ฃ^3 )/(1 โ ๐ฃ^4 )โ3โซ1โใ๐ฃ/(1 โใ ๐ฃใ^4 ) ๐๐ฃใใ Put ๐^๐โ๐=๐ Diff. w.r.t. ๐ฃ ๐/๐๐ฃ (๐ฃ^4โ1)=๐๐ก/๐๐ฃ 4๐ฃ^3=๐๐ก/๐๐ฃ ๐๐ฃ=๐๐ก/(4๐ฃ^3 ) Put ๐=๐^๐ Diff. w.r.t. ๐ฃ ๐๐/๐๐ฃ=2๐ฃ ๐๐/2๐ฃ=๐๐ฃ ๐ฐ =โซ1โใ๐^๐/(โ๐) ๐ ๐/(๐๐^๐ ) โ๐โซ1โใ๐/(๐ โ ๐^๐ ) ๐ ๐/๐๐ใใ ๐ผ =โ1/4 โซ1โใ ๐๐ก/๐กโ3/2 โซ1โใ ๐๐/(1 โ ๐^2 )ใใ ๐ผ =โ1/4 โซ1โใ ๐๐ก/๐ก+3/2 โซ1โใ ๐๐/((๐^2 โ 1^2 ) )ใใ ๐ฐ = (โ๐)/( ๐) ๐ฅ๐จ๐ โก๐+๐/๐ ร ๐/(๐(๐)) ๐๐๐((๐ โ ๐)/(๐ + ๐)) Putting t = ๐ฃ^4 โ 1 and p = v2 I = (โ๐)/๐ ๐ฅ๐จ๐ โกใ(๐^๐โ๐) ใ+๐/๐ ๐ฅ๐จ๐ โกใ((๐^๐ โ ๐))/((๐^๐ + ๐))ใ I = 1/4 [โlogโกใ(๐ฃ^4โ1)+3 ๐๐๐ ((๐ฃ^2 โ 1))/((๐ฃ^2 + 1))ใ ] I = 1/4 [โ๐๐จ๐ โกใ(๐^๐โ๐)+ ๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ใ ] I = 1/4 [๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ร 1/((๐ฃ^4 โ 1))] I = 1/4 [๐๐๐โกใ๐/((๐^๐โ๐) )+ ๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ใ ] I = 1/4 ๐๐๐ ๐/((๐^๐ โ ๐))ร(๐^๐ โ ๐)^๐/(๐^๐ + ๐)^๐ I = 1/4 ๐๐๐ 1/((๐ฃ^2 โ 1)(๐ฃ^2 + 1))ร(๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 I = 1/4 ๐๐๐ (๐ฃ^2 โ 1)^2/(๐ฃ^2 + 1)^4 I = ๐/๐ ๐ฅ๐จ๐ โกใ(((๐^๐ โ ๐))/(๐^๐ + ๐)^๐ )^๐ ใ I = 1/4 ร 2 logโกใ((๐ฃ^2 โ 1))/(๐ฃ^2 + 1)^2 ใ I = ๐/๐ ๐๐๐โกใ((๐^๐ โ ๐))/(๐^๐ + ๐)^๐ ใ Putting back v = ๐ฆ/๐ฅ I = 1/2 logโกใ(((๐ฆ/๐ฅ)^2 โ 1))/((๐ฆ/๐ฅ)^2 + 1)^2 ใ I = 1/2 log (((๐ฆ^2 โ ๐ฅ2)/๐ฅ^2 )/((๐ฆ^2 + ๐ฅ2)/๐ฅ^2 )^2 ) I = ๐/๐ log [(๐๐(๐^๐ โ ๐^๐))/(๐^๐ + ๐^๐ )^๐ ] Substituting value of I in (2) I = log |x| + c ๐/๐ log โ(๐๐(๐๐ โ ๐๐))/((๐๐ + ๐๐))โ = log |x| + c log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = 2 log |x| + 2c log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = log |x|2 + log c1 log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = log |x|2 + log c1 log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/(๐ฅ2 + ๐ฆ2)^2 โ = log c1|x|2 Cancelling log (๐๐(๐๐ โ ๐๐))/(๐๐ + ๐๐)^๐ = c1 x2 x2 (y2 โ x2) = c1 x2 (x2 + y2)2 Cancelling x2 from both sides y2 โ x2 = c1 (x2 + y2)2 x2 โ y2 = c2 (x2 + y2)2 Hence proved