Answer to Question #151328 in Differential Equations for Ashweta Padhan

Question #151328
Find the integral of the Pfaffian differential equation
(1+yz)dx+x(z-x)dy-(1+xy)dz=0
1
Expert's answer
2020-12-17T19:02:16-0500

The necessary and sufficient conditionfor iintegrability is

"R(\\frac{\\partial P}{\\partial y}-\\frac{\\partial Q}{\\partial x})+P(\\frac{\\partial Q}{\\partial z}-\\frac{\\partial R}{\\partial y})+Q(\\frac{\\partial R}{\\partial x}-\\frac{\\partial P}{\\partial z})=0"

"-(1+xy)(z-(z-2x))+(1+yz)(x+x)+x(z-x)(-y-y)="

"=-2x-2x^2y+2x+2xyz-2xyz+2x^2y=0"

Thus, the given equation is integrable.


"(1+yx)dz\u2212(1+yx)dx\u2212x(z\u2212x)dy\u2212y(z\u2212x)dx=0"

"(1+yx)d(z\u2212x)\u2212(xdy+ydx)(z\u2212x)=0"

"(1+yx)d(z\u2212x)\u2212d(1+xy)(z\u2212x)=0"

"\\frac{d(z-x)}{z-x}-\\frac{d(1+xy)}{1+xy}=0"

"ln(z\u2212x)\u2212ln(1+xy)=lnc_1\n\u200b"

"\\frac{z-x}{1+xy}=c_1"


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