Answer to Question #159084 in Differential Equations for YUVASRI RAMAKRISHNAN

Question

Verify that the total differential equation(1+yz)dx+x(z-x)dy-(1+xy)dz=0 is integrable and hence find its integral.

Answer 1

Necessary and sufficient condition for the total differential equation $Pdx+Qdy+Rdz=0$

to be integrable is $P(∂R/∂y-∂Q/∂z)-Q(∂R/∂x-∂P/∂z)+ +R(∂Q/∂x-∂P/∂y)=0$

$P=1+yz$

$∂P/∂z=y$

$∂P/∂y=z$

$Q=x(z-x)$

$∂Q/∂x=z-2x$

$∂Q/∂z=x$

$R=-(1+xy)$

$∂R/∂x=-y$

$∂R/∂y=-x$

According to the given equation

$(1+yz)(-x-x)-x(z-x)(-y-y)+(-1-xy)(z-2x-z)=\\ =-2x-2xyz+2xyz-2x^2 y+2x+2x^2 y=0$

So it is integrable.

 

Let $y=const$

$(1+yz)dx-(1+xy)dz=0$

Integrate

$x+xyz-z-xyz=f(y)$

$x-z=f(y)$

Use the equation

$(1+yz)dx+f' (y)dy-(1+xy)dz=0$

$f' (y)=x(z-x)$

$f' (y)=-f(y)x$

$ln⁡〖f(y)〗=-xy+C$

$f(y)=exp(-xy+C)$

Answer: $x-z=exp(-xy+C)$