dy dy dtdxdt dxdyEx3: Use the chain rule to express-in terndxt2a) y =and t = v2x + 1%3Dt2+11b) y =and x = V4t +1t2+1c) y = (1and x =1at t = 2t+1,t2d) y = 1– 1and t =1-xat x = 2Sol:t2a) y :2t(t2+1)-2t.t2dy2t%3D%3D%3Dt2+1dt(t2+1)2(t2+1)2=!:(2x + 1) 2 = Ei1dtt = (2x + 1)2 =dx1%3D%3D2V2x+1dydy%3Ddt2/2x+12t1%3D%3Ddx(2x+1)+1) VZ=dtdx(t2+1)2 V2x+1b) y = (t2 + 1)-1=dy2t-2t(t2 + 1)-2dt(t2+1)2

Name: dy dy dtdxdt dxdyEx3: Use the chain rule to express-in terndxt2a) y =
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Description: dy dy dtdxdt dxdyEx3: Use the chain rule to express-in terndxt2a) y =and t = v2x + 1%3Dt2+11b) y =and x = V4t +1t2+1c) y = (1and x =1at t = 2t+1,t2d) y = 1– 1and t =1-xat x = 2Sol:t2a) y :2t(t2+1)-2t.t2dy2t%3D%3D%3Dt2+1dt(t2+1)2(t2+1)2=!:(2x + 1) 2

Given rule of derivative dydx=dydt×dtdx

Math

Advanced Math

dy dy dt * dx dt dx Ex3: Use the chain rule to express: in terz dx dy t2 a) y = and t = v2x +1 %3D %3D t2+1 1 b) y = 241 and x = V4t + 1 c) y = ( 1 x 3D t2 and -1 at t = 2 %3D t+1. 1 d) y = 1- 1 and t = 1-x at x = 2

dy dy dt * dx dt dx Ex3: Use the chain rule to express: in terz dx dy t2 a) y = and t = v2x +1 %3D %3D t2+1 1 b) y = 241 and x = V4t + 1 c) y = ( 1 x 3D t2 and -1 at t = 2 %3D t+1. 1 d) y = 1- 1 and t = 1-x at x = 2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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