Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 10 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. (1+๐^(๐ฅ/๐ฆ) )๐๐ฅ+๐^(๐ฅ/๐ฆ) (1โ๐ฅ/๐ฆ)๐๐ฆ=0 Step 1: Find ๐๐ฅ/๐๐ฆ (1+๐^(๐ฅ/๐ฆ) )๐๐ฅ+๐^(๐ฅ/๐ฆ) (1โ๐ฅ/๐ฆ)๐๐ฆ = 0 (1+๐^(๐ฅ/๐ฆ) ) dx = โ๐^(๐ฅ/๐ฆ) (1โ๐ฅ/๐ฆ)๐๐ฆ ๐๐/๐๐ = (โ๐^(๐/๐) (๐ โ ๐/๐) )/(๐ + ๐^(๐/๐) ) Since the equation is in the form ๐ฅ/๐ฆ , we will take ๐๐ฅ/๐๐ฆ Instead of ๐๐ฆ/๐๐ฅ Step 2: Put ๐๐ฅ/๐๐ฆ = F(x, y) and find F(๐x, ๐y) F(x, y) = (โ๐^(๐ฅ/๐ฆ) (1 โ ๐ฅ/๐ฆ) )/(1 + ๐^(๐ฅ/๐ฆ) ) F(๐x, ๐y) = (โ๐^(๐๐ฅ/๐๐ฆ) (1 โ ๐๐ฅ/๐๐ฆ) )/(1 + ๐^(๐๐ฅ/๐๐ฆ) ) = (โ๐^(๐ฅ/๐ฆ) (1 โ ๐ฅ/๐ฆ) )/(1 + ๐^(๐ฅ/๐ฆ) ) = ๐น(๐ฅ, ๐ฆ) Thus, F(x, y) is a homogenous equation function of order zero Therefore ๐๐ฆ/๐๐ฅ is a homogenous differential equation Step 3: Solving ๐๐ฅ/๐๐ฆ by putting ๐ฅ=๐ฃ๐ฆ ๐๐ฅ/๐๐ฆ= (โ๐^(๐ฅ/๐ฆ) (1 โ ๐ฅ/๐ฆ) )/(1 + ๐^(๐ฅ/๐ฆ) ) Put ๐=๐๐ Diff. w.r.t. ๐ฆ ๐๐ฅ/๐๐ฆ=๐/๐๐ฆ (๐ฃ๐ฆ) ๐๐ฅ/๐๐ฆ=๐ฆ . ๐๐ฃ/๐๐ฆ+๐ฃ ๐๐ฆ/๐๐ฆ ๐๐/๐๐=๐ . ๐๐/๐๐+๐ Putting values of ๐๐ฅ/๐๐ฆ and x in (1) ๐๐ฅ/๐๐ฆ=(โ๐^(๐ฅ/๐ฆ) (1 โ ๐ฅ/๐ฆ) )/(1 + ๐^(๐ฅ/๐ฆ) ) ๐+๐ ๐๐/๐๐=(โ๐^๐ (๐ โ ๐))/(๐ + ๐^๐ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ๐^๐ฃ (1 โ ๐ฃ))/(1 + ๐^๐ฃ )โ๐ฃ ๐ฆ ๐๐ฃ/๐๐ฆ=(โ๐^๐ฃ+ ๐ฃ๐^๐ฃ)/(1 + ๐^๐ฃ )โ๐ฃ ๐ฆ ๐๐ฃ/๐๐ฆ=(โ๐^๐ฃ+ ๐ฃ๐^๐ฃ โ ๐ฃ(1 + ๐^๐ฃ ))/(1 + ๐^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ๐^๐ฃ+ ๐ฃ๐^๐ฃ โ ๐ฃ โ ๐ฃ๐^๐ฃ)/(1 + ๐^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ๐^๐ฃโ ๐ฃ)/(1 + ๐^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ(๐^๐ฃ+ ๐ฃ))/(1 + ๐^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ(๐^๐ฃ+ ๐ฃ))/(1 + ๐^๐ฃ ) ใ๐ + ๐ใ^๐/(๐ + ๐^๐ ) ๐๐ = (โ๐๐)/๐ Integrating both sides โซ1โใใ1 + ๐ใ^๐ฃ/(๐ฃ + ๐^๐ฃ ) ๐๐ฃ" " ใ =โซ1โ(โ๐๐ฆ)/๐ฆ โซ1โใใ๐ + ๐ใ^๐/(๐ + ๐^๐ ) ๐๐ใ=โ๐ฅ๐จ๐ โกใ|๐|ใ+๐๐๐โก๐ Put v + ev = t (1 + ev) dv = dt Thus, our equation becomes โซ1โ๐๐ก/๐ก=โlogโกใ|๐ฆ|ใ+logโก๐ ๐ฅ๐จ๐ โกใ|๐|ใ=โ๐๐๐โกใ|๐|ใ+๐๐๐โก๐ Putting back value of t = v + ev logโกใ|๐ฃ+๐^๐ฃ |ใ=โlogโกใ|๐ฆ|ใ+logโก๐ logโกใ|๐ฃ+๐^๐ฃ |ใ+logโกใ|๐ฆ|ใ=logโก๐ logโก(|๐ฃ+๐^๐ฃ |ร|๐ฆ|)=logโก๐ logโก((๐ฃ+๐^๐ฃ )ร๐ฆ)=logโก๐ ๐๐๐โก(๐๐+๐^๐ ๐)=๐๐๐โก๐ Putting back value of v = ๐ฅ/๐ฆ logโก(๐ฅ/๐ฆร๐ฆ+๐^(๐ฅ/๐ฆ) ๐ฆ)=logโก๐ ๐๐๐โก(๐ฅ+๐^(๐ฅ/๐ฆ) ๐ฆ)=๐๐๐โก๐ Canceling log ๐+๐๐^(๐/๐)=๐ช