Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 9 In each of the Exercise 1 to 10, show that the given differential equation is homogeneous and solve each of them. ๐ฆ ๐๐ฅ+๐ฅ๐๐๐ (๐ฆ/๐ฅ)๐๐ฆโ2๐ฅ ๐๐ฆ=0 Step 1: Find ๐๐ฆ/๐๐ฅ ๐ฆ ๐๐ฅ+๐ฅ๐๐๐ (๐ฆ/๐ฅ)๐๐ฆโ2๐ฅ ๐๐ฆ=0 dy [๐ฅ logโกใ(๐ฆ/๐ฅ)โ2๐ฅ ใ ] = โ y dx ๐๐ฆ/๐๐ฅ = (โ๐ฆ)/(๐ฅ logโกใ(๐ฆ/๐ฅ) โ 2๐ฅ ใ ) ๐๐ฆ/๐๐ฅ = (โ๐ฆ)/(โ๐ฅ(2 โ logโก(๐ฆ/๐ฅ) ) ) ๐๐/๐๐ = (๐/๐)/(๐ โ ๐๐๐โก(๐/๐) ) Step 2: Putting F(x , y) = ๐๐ฆ/๐๐ฅ and finding F(๐x, ๐y) F(x, y) = (๐ฆ/๐ฅ)/(2 โ logโก(๐ฆ/๐ฅ) ) ๐น(๐๐ฅ,๐๐ฆ) = (๐๐ฆ/๐๐ฅ)/(2 โ logโก(๐๐ฆ/๐๐ฅ) ) = (๐ฆ/๐ฅ)/(2 โ logโก(๐ฆ/๐ฅ) ) = ๐ยฐ [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ฆ/๐๐ฅ by putting y = vx Putting y = vx Diff w.r.t.x ๐๐ฆ/๐๐ฅ = x ๐๐ฃ/๐๐ฅ + v ๐๐ฅ/๐๐ฅ ๐๐/๐๐ = x ๐๐/๐๐ + v Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ = (๐ฆ/๐ฅ)/(2 โ logโก(๐ฆ/๐ฅ) ) v + (๐ ๐๐)/๐๐ = (๐๐/๐)/(๐ โ ๐๐๐(๐๐/๐) ) v + (๐ฅ ๐๐ฃ)/๐๐ฅ = ๐ฃ/(2 โ logโก๐ฃ ) (๐ฅ ๐๐ฃ)/๐๐ฅ = ๐ฃ/(2 โ logโก๐ฃ ) โ v (๐ฅ ๐๐ฃ)/๐๐ฅ = (๐ฃ โ 2๐ฃ + ๐ฃ logโก๐ฃ)/(2 โ logโก๐ฃ ) (๐ฅ ๐๐ฃ)/๐๐ฅ = (๐ฃ logโก๐ฃ โ ๐ฃ)/(2 โใ logใโก๐ฃ ) (๐ โ ๐๐๐โก๐)/(๐ ๐๐๐โก๐ ) dv = ๐๐/๐ Integrating both sides โซ1โ(2 โ logโก๐ฃ)/(๐ฃ logโก๐ฃ โ ๐ฃ) dv = โซ1โ๐๐ฅ/๐ฅ โซ1โ(2 โ logโก๐ฃ)/(โ๐ฃ ใ(1 โ logใโก๐ฃ)) dv = log |x| + log c โซ1โ(1 + 1 โ logโก๐ฃ)/(โ๐ฃ ใ(1 โ logใโก๐ฃ)) ๐๐ฃ= log |x| + log c โซ1โ1/((โ๐ฃ)(1 โ logโกใ๐ฃ)ใ ) ๐๐ฃ โโซ1โ1/๐ฃ ๐๐ฃ = log |x| + log c โซ1โ1/(๐ฃ(logโก๐ฃ โ 1)) ๐๐ฃ โ โซ1โ1/๐ฃ ๐๐ฃ = log |x| + log c โซ1โ๐๐/(๐(๐๐๐โกใ๐ โ ๐)ใ ) โ log |v| = log |x| + log c Put t = log v โ 1 dt = 1/๐ฃ dv So, our equation becomes โซ1โ๐๐ก/๐ก โ log v = log x + log c log t โ log v = log x + log c Putting value of t log (log v โ 1) โlog v + = log x + log c log (log v โ 1) = log x + log c + log v log (log v โ 1) = log Cxv Putting value of v = ๐ฆ/๐ฅ log ("log " ๐ฆ/๐ฅโ1)=logโกใ๐ฅ ๐ ๐ฆ/๐ฅใ log ("log " ๐/๐โ๐)=๐๐๐โกใ ๐๐ใ log ๐ฆ/๐ฅ โ 1 = cy cy = log |๐/๐| โ 1