Answer to Question #199550 in Differential Equations for Rajkumar

Question

Find the surface which is orthogonal to the one parameter system z = c xy( x² + y² )

which passes through the hyperbola , x² - y² = a² , z =0.

Answer 1

$$𝑓(𝑥, 𝑦, 𝑧) =\dfrac{z}{xy(x^2+y^2)}=c$$

$$f_x=-\dfrac{z(3x^2+y^2)}{x^2y(x^2+y^2)^2}$$

$$f_y=-\dfrac{z(3y^2+x^2)}{xy^2(x^2+y^2)^2}$$

$$f_z=\dfrac{1}{xy(x^2+y^2)}$$

Let $z(x,y)$ is a surface orthogonal to the given system. Then:

$$(f_x, f_y, f_z)\cdot(z_x, z_y, -1)=0$$

So, we have differential equation:

$$-\dfrac{z(3x^2+y^2)}{x^2y(x^2+y^2)^2}z_x-\dfrac{z(3y^2+x^2)}{xy^2(x^2+y^2)^2}z_y$$

$$-\dfrac{1}{xy(x^2+y^2)}=0$$

Divide by $-\dfrac{z}{xy(x^2+y^2)}$

$$\dfrac{3x^2+y^2}{x(x^2+y^2)}z_x+\dfrac{3y^2+x^2}{y(x^2+y^2)}z_y+\dfrac{1}{z}=0$$

The auxiliary equations:

$$\dfrac{x(x^2+y^2)}{3x^2+y^2}dx=\dfrac{y(x^2+y^2)}{3y^2+x^2}dy=-zdz$$

Adding first and second and equating to third:

$$\dfrac{(x^2+y^2)(xdx+ydy)}{4x^2+4y^2}=-zdz$$

$$\dfrac{(xdx+ydy)}{4}=-zdz$$

giving 

$$x^2+y^2+4z^2=c_1$$

Subtracting the second equation from the first one:

$$\dfrac{(x^2+y^2)(xdx-ydy)}{2(x^2-y^2)}$$

$$\dfrac{(c_1-4z^2)(xdx-ydy)}{2(x^2-y^2)}=-zdz$$

$$x^2-y^2=c_2\sqrt{c_1-4z^2}$$

$$c_2=\dfrac{x^2-y^2}{\sqrt{c_1-4z^2}}$$

Using the given conditions $x^2-y^2=a, z=0:$

$$a^2=c_1c_2^2$$

Then

$$a^2=(x^2+y^2+4z^2)\dfrac{(x^2-y^2)^2}{x^2+y^2}$$

$$x^2+y^2+4z^2=\dfrac{a^2(x^2+y^2)}{(x^2-y^2)^2}$$

$$z=\sqrt{\dfrac{a^2(x^2+y^2)}{4(x^2-y^2)^2}-\dfrac{x^2+y^2}{4}}$$