# Ex 9.5, 3

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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. 𝑥−𝑦𝑑𝑥− 𝑥+𝑦𝑑𝑥=0 Step 1 Find 𝑑𝑦𝑑𝑥 (x − y) dy − (x + y) dx = 0 (x − y) dy = (x + y) dx Step 2: Put 𝑑𝑦𝑑𝑥 = F(x, y) and find out F(𝜆x, 𝜆y) F(x, y) = 𝑥 + 𝑦𝑥 − 𝑦 F(𝜆x, 𝜆y) = 𝜆𝑥 + 𝜆𝑦𝜆𝑥−𝜆𝑦 = 𝜆(𝑥 + 𝑦)𝜆 (𝑥 − 𝑦) = 𝑥 + 𝑦𝑥 − 𝑦 = F(x, y) ∴ F(𝜆x, 𝜆y) = 𝜆0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦𝑑𝑥 is a homogenous differential equation. Step 3. Solving 𝑑𝑦𝑑𝑥 by putting y = vx Put y = vx. differentiating w.r.t.x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥+ 𝑣𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑥 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥 = 𝑥 + 𝑦𝑥−𝑦 x 𝑑𝑣𝑑𝑥 + v = 𝑥 + 𝑥𝑣𝑥 − 𝑥𝑣 x 𝑑𝑣𝑑𝑥 + v = 𝑥 (1 + 𝑣)𝑥(1 − 𝑣) x 𝑑𝑣𝑑𝑥 + v = (1 + 𝑣)(1 − 𝑣) x 𝑑𝑣𝑑𝑥 = (1 + 𝑣)(1 − 𝑣) − v x 𝑑𝑣𝑑𝑥 = 1 + 𝑣 − 𝑣(1 − 𝑣)1 − 𝑣 x 𝑑𝑣𝑑𝑥 = 1 + 𝑣 − 𝑣 + 𝑣21 − 𝑣 x 𝑑𝑣𝑑𝑥 = 1 + 𝑣21 − 𝑣 1 − 𝑣𝑑𝑣1 + 𝑣2 = 𝑑𝑥𝑥 Integrating both sides 1 − 𝑣1 + 𝑣2𝑑𝑣= 𝑑𝑥𝑥 tan−1 v − 𝑣1 + 𝑣2 = log 𝑥+𝑐 Let I = 𝑣1 + 𝑣2 dv Put t = 1 + 𝑣2 Diff w.r.t. v 𝑑𝑑𝑥(1 + v2) = 𝑑𝑡𝑑𝑣 2v = 𝑑𝑡𝑑𝑣 dv = 𝑑𝑡2𝑣 Therefore I = 𝑣1 + 𝑣2 dv = 𝑑𝑡2𝑡 = 12𝑙𝑜𝑔 𝑡+𝑐 Putting t = 1 + v2 = 12𝑙𝑜𝑔 1+ 𝑣2 + c Putting value of I in (2) tan−1 v − 12 log 1+𝑣2 = log 𝑥 + c tan−1 v = 12 log 1+𝑣2 + log 𝑥 + c tan−1 v = 12 log 1+𝑣2 + 22 log 𝑥 + c tan−1 v = 12 log 1+𝑣2 + 2 log 𝑥 + c tan−1 v = 12 log 1+𝑣2. 𝑥2 + c Putting v = 𝑦𝑥 tan−1 𝑦𝑥= 12 log 1+ 𝑦𝑥2× 𝑥2+c tan−1 𝑦𝑥= 12 log 𝑥2 + 𝑦2 𝑥2× 𝑥2+c tan−1 𝑦𝑥= 12 log 𝑥2+ 𝑦2+c tan−1 𝒚𝒙= 𝟏𝟐 log 𝒙𝟐+ 𝒚𝟐+𝐜 is the required solution

Chapter 9 Class 12 Differential Equations

Serial order wise

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