Question about Euler’s Equations when Auxiliary Conditions are Imposed

[sams]

In the Classical Dynamics of Particles and Systems book, 5th Edition, by Stephen T. Thornton and Jerry B. Marion, page 220, the author derived Equation (6.67) from Equation (6.66) which is the following:

Equation (6.67):

$$\left(\frac{\partial f}{\partial y} − \ \frac{d}{dx}\frac{\partial f}{\partial y^′}\right)\left(\frac{\partial g}{\partial y}\right)^{−1} = \ \left(\frac{\partial f}{\partial z} − \ \frac{d}{dx}\frac{\partial f}{\partial z^′} \right) \left(\frac{\partial g}{\partial z} \right)^{−1}$$

Because $y$ and $z$ are both functions of $x$, the two sides of Equation (6.67) may be set equal to a function of $x$ which we write as $\lambda(x)$:

Equations (6.68):
$$\frac{\partial f}{\partial y} − \frac{d}{dx}\frac{\partial f}{\partial y^′} + \lambda(x)\frac{\partial g}{\partial y} = 0$$ $$\frac{\partial f}{\partial z} − \frac{d}{dx}\frac{\partial f}{\partial z^′} + \lambda(x)\frac{\partial g}{\partial z} = 0$$

where $\lambda(x)$ is the Lagrange undetermined multiplier.
How did the author deduce Equations (6.68) from Equation (6.67)?

Any help is much appreciated. Thank you so much.

 

[haruspex]

In the Classical Dynamics of Particles and Systems book, 5th Edition, by Stephen T. Thornton and Jerry B. Marion, page 220, the author derived Equation (6.67) from Equation (6.66) which is the following:

Equation (6.67):

$$\left(\frac{\partial f}{\partial y} − \ \frac{d}{dx}\frac{\partial f}{\partial y^′}\right)\left(\frac{\partial g}{\partial y}\right)^{−1} = \ \left(\frac{\partial f}{\partial z} − \ \frac{d}{dx}\frac{\partial f}{\partial z^′} \right) \left(\frac{\partial g}{\partial z} \right)^{−1}$$

Because $y$ and $z$ are both functions of $x$, the two sides of Equation (6.67) may be set equal to a function of $x$ which we write as $\lambda(x)$:

Equations (6.68):
$$\frac{\partial f}{\partial y} − \frac{d}{dx}\frac{\partial f}{\partial y^′} + \lambda(x)\frac{\partial g}{\partial y} = 0$$ $$\frac{\partial f}{\partial z} − \frac{d}{dx}\frac{\partial f}{\partial z^′} + \lambda(x)\frac{\partial g}{\partial z} = 0$$

where $\lambda(x)$ is the Lagrange undetermined multiplier.
How did the author deduce Equations (6.68) from Equation (6.67)?

Any help is much appreciated. Thank you so much.

Almost exactly as described: e.g. take the LHS of 6.67 and write that it equals -λ(x). Multiply through by $\frac{\partial g}{\partial y}$.