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Question Number 31485 by mondodotto@gmail.com last updated on 09/Mar/18

Answered by ajfour last updated on 09/Mar/18

(i)   let    y−2 = t(x−3)  ⇒    (dy/dx)=(x−3)(dt/dx)+t  Then  diff. eq. becomes          (x−3)(dt/dx)+t = ((t(x−3))/((t+1)(x−3)))  ⇒  (x−3)(dt/dx) +t =(t/(t+1))  or    (x−3)(dt/dx) = (t/(t+1))−t           ∫(((t+1)dt)/t^2 ) = −∫(dx/(x−3))  ⇒   ln ∣t∣−(1/t) = −ln ∣x−3∣+c  ⇒   ln ∣((y−2)/(x−3))∣+ln ∣x−3∣=((x−3)/(y−2))+c   ⇒  ln ∣y−2∣=((x−3)/(y−2))+c .

(i)lety2=t(x3)dydx=(x3)dtdx+tThendiff.eq.becomes(x3)dtdx+t=t(x3)(t+1)(x3)(x3)dtdx+t=tt+1or(x3)dtdx=tt+1t(t+1)dtt2=dxx3lnt1t=lnx3+clny2x3+lnx3∣=x3y2+clny2∣=x3y2+c.

Answered by ajfour last updated on 09/Mar/18

(ii)   let   2x+y+1 =t  ⇒               (dy/dx)=(dt/dx)−2  Differential eq. becomes              (dt/dx)−2 = ((t−3)/t)  ⇒     (dt/dx)=((3t−3)/t)  or       ∫(t/(t−1))dt = 3∫dx   ⇒    ∫dt+∫(dt/(t−1)) = 3x+c  or    t+ln ∣t−1∣=3x+c  ⇒ 2x+y+1+ln ∣2x+y∣=3x+c .

(ii)let2x+y+1=tdydx=dtdx2Differentialeq.becomesdtdx2=t3tdtdx=3t3tortt1dt=3dxdt+dtt1=3x+cort+lnt1∣=3x+c2x+y+1+ln2x+y∣=3x+c.

Commented by mondodotto@gmail.com last updated on 09/Mar/18

thanx a lot.

thanxalot.

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