Change of the dependent variable

Let

$$y = f(x),$$

and suppose at the same time $y$ is a function of $z$, say

$$y = g(z).$$

We may then express $\frac{dy}{dx}$, $\frac{d^2 y}{dx^2}$ etc., in terms of $\frac{dz}{dx}$, $\frac{d^2 z}{dx^2}$, etc., as follows

In general, $z$ is a function of $y$, and since $y$ is a function of $x$, it is evident that $z$ is a function of $x$. Hence by XXV of §5.1, we have

$$\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \psi'(z) \frac{dz}{dx}.$$

Also $\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( g'(z) \frac{dz}{dx} \right) = \frac{dz}{dx} \frac{d}{dx} g'(z) + g'(z) \frac{d^2 z}{dx^2}$. But $\frac{d}{dx} g'(z) = \frac{d}{dz} g'(z) \frac{dz}{dx} = g''(z) \frac{dz}{dx}$. Therefore,

$$\frac{d^2 y}{dx^2} = g''(z) \left( \frac{dz}{dx} \right)^2 + g'(z) \frac{d^2 z}{dx^2}.$$

Similarly for higher derivatives. This transformation is called changing the dependent variable from $y$ to $z$, the independent variable remaining $x$ throughout. We will now illustrate this process by means of an example.

Example 11.2.1   Having given the equation

$$\frac{d^2 y}{dx^2} = 1 + \frac{2( 1 + y)}{1 + y^2} \left( \frac{dy}{dx} \right)^2,$$

change the dependent variable from $y$ to $z$ by means of the relation

$$y = \tan\, z.$$

Solution. From the above,

$$\frac{dy}{dx} = \sec^2 (z) \frac{dz}{dx}, \frac{d^2 y}{dx^2} = ... ...z) \frac{d^2 z}{dx^2} + 2 \sec^2 (z) \tan (z) \left( \frac{dz}{dx} \right)^2,$$

Substituting,

$$\sec^2 (z) \frac{d^2 z}{dx^2} + 2 \sec^2 (z) \tan (z) \left( \fr... ...+ \frac{2(1 + \tan z)}{1 + \tan^2 z} \left( \sec^2 z \frac{dz}{dx} \right)^2,$$

and reducing, we get $\frac{d^2 z}{dx^2} - 2 \left( \frac{dz}{dx} \right)^2 = \cos^2 z$.