Answer to Question #323276 in Differential Equations for Sneka

Question

45) Find the integral surface of the equation x(y ^ 2 + z) * p - y(x ^ 2 + z) * q = (x ^ 2 - y ^ 2) * z whic the straight line x + y = 0, z = 1

Answer 1

$x(y ^ 2 + z)\frac{\partial z}{dx}- y(x ^ 2 + z)\frac{\partial z}{dy}=(x ^ 2 - y ^ 2) z$

$x + y = 0, z = 1$

The auxiliary equations is:

$\frac{dx}{x(y ^ 2 + z)}=\frac{dy}{- y(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}$

A first characteristic equation comes from

$\frac{xdx+ydy}{x^2(y ^ 2 + z)- y^2(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}$

$\frac{xdx+ydy}{(x ^ 2 - y ^ 2) z}=\frac{dz}{(x ^ 2 - y ^ 2) z}$

$xdx+ydy=dz$

$\frac{x^2+y^2}{2}=z+\frac{C_1}{2}$

$C_1=x^2+y^2-2z$

A second characteristic equation comes from

$\frac{ydx+xdy}{xy(y ^ 2 + z)- xy(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}$

$\frac{ydx+xdy}{-xy(x ^ 2 - y ^ 2) }=\frac{dz}{(x ^ 2 - y ^ 2) z}$

$\frac{d(xy)}{-xy}=\frac{dz}{ z}$

$-\ln{|xy|}+\ln{C_2}=\ln {|z|}$

$\frac{C_2}{xy}=z$

$C_2=xyz$

General solution of the PDE on the form of implicit equation:

$\Phi(C_1,C_2)=0$.

$C_1=x^2+y^2-2z=x^2+y^2+2xy-2xy-2z=$$(x+y)^2-2xy-2z$ (1)

$C_2=xyz$ (2)

Substitute (2) into (1):

$C_1=(x+y)^2-2\frac{C_2}{z}-2z$

Use $x + y = 0, z = 1$ to exclude $C_1$ and $C_2$ :

$C_1=(0)^2-2\frac{C_2}{1}-2\cdot1=-2C_2-2$

$-2C_2-2=x^2+y^2-2z$

$-2xyz-2=x^2+y^2-2z$

$z=\frac{x^2+y^2}{2}+xyz+1$

Answer: $z=\frac{x^2+y^2}{2}+xyz+1$ .