Solve the linear differential equation [1+ytan(xy)]dx+xtn(xy)dy=0

Linear Equations of Order one
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14 Answers

P_=1+Y•TAN(xy)
by Level 12 User (101k points)
Q_=x•TAN(xy)
by Level 12 User (101k points)
}→æp/æy=æQ/æx
by Level 12 User (101k points)
=TAN(xy)+xy.SEC^2(xy)
by Level 12 User (101k points)
Therefore I.F=1
by Level 12 User (101k points)
And du_=Pdx+Qdy=0
by Level 12 User (101k points)
→u(x,y)=c
by Level 12 User (101k points)
æu/æy=Q
by Level 12 User (101k points)
U=w(x)+x.{TAN(xy)dy
by Level 12 User (101k points)
→u=w(x)+luSEC(xy)
by Level 12 User (101k points)
æu/æx=P_=w'(x)-æ/æx[luSEC(xy)]
by Level 12 User (101k points)
→w(x)=x
by Level 12 User (101k points)
Therefore x+luSEC(xy)=c ~upper
by Level 12 User (101k points)
→Y=1/x.ARCOS(ce^2)
by Level 12 User (101k points)

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