# Ex 9.5, 3 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 3 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. + =0 Step 1 Find (x y) dy (x + y) dx = 0 (x y) dy = (x + y) dx Step 2: Put = F(x, y) and find out F( x, y) F(x, y) = + F( x, y) = + = ( + ) ( ) = + = F(x, y) F( x, y) = 0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, is a homogenous differential equation. Step 3. Solving by putting y = vx Put y = vx. differentiating w.r.t.x = x + = + v Putting value of and y = vx in (1) = + x + v = + x + v = (1 + ) (1 ) x + v = (1 + ) (1 ) x = (1 + ) (1 ) v x = 1 + (1 ) 1 x = 1 + + 2 1 x = 1 + 2 1 1 1 + 2 = Integrating both sides 1 1 + 2 = tan 1 v 1 + 2 = log + Let I = 1 + 2 dv Put t = 1 + 2 Diff w.r.t. v (1 + v2) = 2v = dv = 2 Therefore I = 1 + 2 dv = 2 = 1 2 + Putting t = 1 + v2 = 1 2 1+ 2 + c Putting value of I in (2) tan 1 v 1 2 log 1+ 2 = log + c tan 1 v = 1 2 log 1+ 2 + log + c tan 1 v = 1 2 log 1+ 2 + 2 2 log + c tan 1 v = 1 2 log 1+ 2 + 2 log + c tan 1 v = 1 2 log 1+ 2 . 2 + c Putting v = tan 1 = 1 2 log 1+ 2 2 +c tan 1 = 1 2 log 2 + 2 2 2 +c tan 1 = 1 2 log 2 + 2 +c tan 1 = log + + is the required solution

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.