Answer to Question #266936 in Differential Equations for Devil

Question

Xyp+y^2q+2x-xyz

Answer 1

Solution;

$xyp+y^2q+2x^2-xyz$

Rewrite as;

$xyp+y^2q=xyz-2x^2$

Here;

$P=xy$

$Q=y^2$

$R=xyz-2x^2$

Axillary equations are;

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

$\frac{dx}{xy}=\frac{dy}{y^2}=\frac{dz}{xyz-2x^2}$ ......(1)

Taking the first two fractions in (1)

$\frac{dx}{xy}=\frac{dy}{y^2}$

Simplify;

$\frac{dx}{x}=\frac{dy}{y}$

Integrating respectively;

$log(x)=log(y)+log(c_1)$

Therefore;

$log(\frac xy)=log(c_1)\implies\frac xy=_-^+c_1=k$ .....(2)

Taking the last two equations in (1);

$\frac{dy}{y^2}=\frac{dz}{xyz-2x^2}$

Substituting (2);

$\frac{dy}{y^2}=\frac{dz}{ky^2z-2k^2y^2}$

Simplify as;

$kdy=\frac{1}{z-2k}dz$

Integrating ;

$ky=log(z-2k)+c_2$

Using (2) as substitution;

$x-log(z-\frac{2x}{y})=c_2$ ....(3)

The general solution is given by (2) and (3);

$(\frac xy,x-log(z-\frac{2x}{y})=0$