Finding Orthogonal Trajectories (differential equations)

[Westlife]

Homework Statement

Find Orthogonal Trajectories of $\frac{x^2}{a}-\frac{y^2}{a-1}=1$
Hint
Substitute a new independent variable w
$x^2=w$
and an new dependent variable z
$y^2=z$

Homework Equations

The Attempt at a Solution

substituting $x$ and $y$ I get
$$\frac{w}{a}-\frac{z}{a-1}=1\quad /a(a-1)$$
$$w(a-1)-za=a(a-1)\quad /'$$
$$w'(a-1)-z'a=0$$

Here I wanted to figure out what $w'$and $z'$ are and here is where I get confused. How do I get them?
since w is now the new independent variable is $w'= \frac{dw}{dw}=1$ and $z'=\frac{dz}{dw}$ or does mean mean something else ?

 

[SammyS]

Homework Statement

Find Orthogonal Trajectories of $\frac{x^2}{a}-\frac{y^2}{a-1}=1$
Hint
Substitute a new independent variable w
$x^2=w$
and an new dependent variable z
$y^2=z$

Homework Equations

The Attempt at a Solution

substituting $x$ and $y$ I get $$\frac{w}{a}-\frac{z}{a-1}=1\quad /a(a-1)$$ $$w(a-1)-za=a(a-1)\quad /'$$ $$w'(a-1)-z'a=0$$

Here I wanted to figure out what $w'$and $z'$ are and here is where I get confused. How do I get them?
since w is now the new independent variable is $w'= \frac{dw}{dw}=1$ and $z'=\frac{dz}{dw}$ or does mean mean something else ?

Hello @Westlife .

Since the new independent variable is $\ w\,,\$ it follows that you take the derivative with respect to $\ w\,.\$ That is to say, take $\ \frac{d}{dw}\,.$

So, yes, you get

$\displaystyle \ (a-1)-a\,z'=0 \text{ , where } z'=\frac{dz}{dw} \,.$​