IVP $x(y-z) z_x + y(z-x) z_y = z(x-y) \; ; \; z=t $ over the initial Curve : $C : x=t, y = 2t/(t^2-1) \; 0 < t < 1$

Question

Exercise :

Solve the IVP : $$\begin{cases} x(y-z) z_x + y(z-x) z_y = z(x-y) \; ; \; z=t \\ C : x=t, y = 2t/(t^2-1) \; \; \; 0 < t < 1 \end{cases}$$

Attempt :

We yield the differential problem :

$$\frac{\mathrm{d}x}{x(y-z)} = \frac{\mathrm{d}y}{y(z-x)} = \frac{\mathrm{d}z}{z(x-y)}$$

and then we'll calculate two integral curves, $z_1, z_2$.

$$\frac{\mathrm{d}x}{x(y-z)} = \frac{\mathrm{d}y}{y(z-x)} \Rightarrow \int y(z-x)\mathrm{d}x = \int x(y-z) \mathrm{d}y$$ $$\implies$$ $$z_1 = \frac{x^2y}{2} + \frac{xy^2}{2} - 2xyz $$

$$\frac{\mathrm{d}x}{x(y-z)} = \frac{\mathrm{d}z}{z(x-y)} \Rightarrow \int z(x-y)\mathrm{d}x = \int x(y-z) \mathrm{d}z$$ $$\implies$$ $$z_2 = \frac{x^2z}{2} + \frac{xz^2}{2} - 2xyz $$

We have the parameters of the initial curve $C$ as the point vector :

$$C : r_C(t) = \bigg(t,\frac{2t}{t^2-1}, t\bigg) \; 0 < t <1$$

and then we yield the system :

$$\begin{cases} Z_1 = \frac{2t^3}{(t^2-1)^2} - \frac{3t^3}{t^2-1} \ Z_2 = t^3 - \frac{4t^3}{t^2-1} \end{cases}

Question - Problem : Now, in this particular step, I have to "get rid" of the variable $t$ by finding a connection between $Z_1, Z_2$, in order to replace the curves back and calculate a general solution. But in this particular case I haven't been able to find a connection between them. Is my solution mistaken at any part ? If not, how to proceed ?

Answer 1

You can do this

$$\frac{\mathrm{d}x}{x(y-z)} = \frac{\mathrm{d}y}{y(z-x)} = \frac{\mathrm{d}z}{z(x-y)}$$ $$\frac{d(x+y)}{z(y-x)} = \frac{\mathrm{d}z}{z(x-y)}$$ $$\int {d(x+y)} = -\int dz$$ $$ \implies x+y+z=c_1$$

---

Edit

As @Lutzl has pointed out it's more simple ( and far better ) to do this $$\frac{y\,dx}{xy(y-z)}=\frac{x\,dy}{xy(z-x)}=\frac{dz}{z(x-y)} $$

$$\implies \frac{y\,dx+x\,dy}{xy(y-x)}=\frac{dz}{z(x-y)}\iff \frac{d(xy)}{xy}+\frac{dz}{z}=0,$$ $$c_2=xyz$$

---

Another way

$$\frac{\mathrm{d}x}{x(y-z)} = \frac{\mathrm{d}y}{y(z-x)} $$

$$y(z-x)dx= {x(y-z)} dy $$ But $$z=c_1-x-y$$ $$y(c_1-x-y-x)dx= {x(y-(c_1-x-y))} dy $$ Rearrange terms $$-(x+y)^2dx+x^2dx+c_1ydx=(x+y)^2dy-y^2dy-c_1xdy$$ $$x^2dx+c_1(ydx+xdy)=(x+y)^2(dy+dx)-y^2dy$$ $$x^2dx+c_1d(xy)=(x+y)^2(d(x+y))-y^2dy$$ Thats easy to integrate $$c_1xy+c_2=x^2y+y^2x$$ $$c_2=xy(y+x-c_1)$$ $$c_2=-xyz$$ $$c_2=xyz$$ If I made no mistakes.......