# Ex 9.5, 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.5, 10 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. 1+ 𝑒 𝑥𝑦𝑑𝑥+ 𝑒 𝑥𝑦 1− 𝑥𝑦𝑑𝑦=0 Step 1 : Find 𝑑𝑥𝑑𝑦 1+ 𝑒 𝑥𝑦𝑑𝑥+ 𝑒 𝑥𝑦 1− 𝑥𝑦𝑑𝑦 = 0 1+ 𝑒 𝑥𝑦 dx = − 𝑒 𝑥𝑦 1− 𝑥𝑦𝑑𝑦 𝑑𝑥𝑑𝑦 = − 𝑒 𝑥𝑦 1 − 𝑥𝑦 1 + 𝑒 𝑥𝑦 Step 2 : Put 𝑑𝑥𝑑𝑦 = F(x, y) and find F(𝜆 x, 𝜆 y) F(x, y) = − 𝑒 𝑥𝑦 1 − 𝑥𝑦 1 + 𝑒 𝑥𝑦 F(𝜆x, 𝜆y) = − 𝑒 𝜆𝑥𝜆𝑦 1 − 𝜆𝑥𝜆𝑦 1 + 𝑒 𝜆𝑥𝜆𝑦 = − 𝑒 𝑥𝑦 1 − 𝑥𝑦 1 + 𝑒 𝑥𝑦= 𝜆° 𝐹(𝑥, 𝑦) Thus, F(x, y) is a homogenous equation function of order zero Therefore 𝑑𝑦𝑑𝑥 is a homogenous differential equation Step 3 : Solving 𝑑𝑦𝑑𝑥 by putting y = vx Putting y = vx Diff w.r.t.x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v 𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥= − 𝑒 𝑥𝑦 1 − 𝑥𝑦 1 + 𝑒 𝑥𝑦 v + y 𝑑𝑣𝑑𝑦 = 𝑒 𝑣𝑦𝑦 1 − 𝑣𝑦𝑦1 + 𝑒 𝑣𝑦𝑦 v + y 𝑑𝑣𝑑𝑦 = 𝑒𝑣 1 − 𝑣1 + 𝑒𝑣 y 𝑑𝑣𝑑𝑦 = 𝑒𝑣 + 𝑣 𝑒𝑣1 + 𝑒𝑣 − v y 𝑑𝑣𝑑𝑦 = − 𝑒𝑣 + 𝑣 𝑒𝑣 − 𝑣 − 𝑣 𝑒𝑣1 + 𝑒𝑣 y 𝑑𝑣𝑑𝑦 = − 𝑣 + 𝑒𝑣1 + 𝑒𝑣 1 + 𝑒𝑣𝑣 + 𝑒𝑣 dv = −𝑑𝑦𝑦 Integrating both sides 1+𝑒𝑣𝑣+ 𝑒𝑣 dv = −𝑑𝑦𝑦 1 + 𝑒𝑣𝑣 + 𝑒𝑣 dv =− log𝑦+ log𝑐 Put v + ev = t (1 + ev) dv = dt Thus, our equation becomes 𝑑𝑡𝑡=− log𝑦+ log𝑐 log𝑡= log𝑦+ log𝑐 Put value of t log (v + ev) = − log y + log c log (v + ev) + log y = log c log y (v + ev) + log c Put value of v = 𝑥𝑦 log y 𝑥𝑦e 𝑥𝑦 log y 𝑥𝑦+ e 𝑥𝑦 = log c y 𝑥𝑦+ e 𝑥𝑦 = c x + y 𝒆 𝒙𝒚 = C

Chapter 9 Class 12 Differential Equations

Serial order wise

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