Question

b) Solve (e'x+e)y'=x-e using the substitution v= e' (this substitution will produce a homogeneous equation).

with clear steps please.

Answer 1

given (e. are 24 y = x-edo By substituting ved in Now ③ in (v.x+v2) I, dv = x-v v de → v(x+u). + d iv x -v Xtv dva 1-v/X tva (It is in Homogeneous form) Nout put z = v=zx + dv. Z + t.dz » Z+3.07 da HZ -Z = 1-7-7- 1+2 1-22-2 HZ

→ (2+1) dz d a (1-22-2) Integrating on both sides Now Here put 1-2z-zhat dt= -2dZ-2 Zdz. at=-21 Z+Daz - 1 Int = lnk+ Ine where is constant ? - It Inic • lot = -2 - 1 Int=Ine-2 cc 2-6 Put & = 1-22-22 in to → 1-27-22.680-2 → 2² +22+(202_120 It is in the form aa²+be+C =0 where the roots are -btJ6²-yac

> z= -2 + J4_4[2652-1) → 7 -2+2 Si-bc)2+1 > z= + √2-452 - we know that z = = --603-3 98. -11212-2012 > y = ln (-2 ta Ja_002) :. Solutions are To ln(-x + x2-202) where is constant