Change of the independent variable

Let $y$ be a function of $x$, and at the same time let $x$ (and hence also $y$) be a function of a new variable $t$. It is required to express

$$\frac{dy}{dx}, \ \ \ \frac{d^2 y}{dx^2},\ \ {\rm etc.},$$

in terms of new derivatives having $t$ as the independent variable. By XXV §5.1, $\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt},$ or

$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }.$$ (11.5)

This is another formulation of the so-called chain rule. Also

$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) ... ...t}{dx} = \frac{ \frac{d}{dt} \left( \frac{dy}{dx} \right) }{ \frac{dx}{dt} }.$$

But differentiating $\frac{dy}{dx}$ with respect to $t$,

$$\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( ... ...dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^2 }.$$

Therefore

$$\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dx^2} }{ \left( \frac{dx}{dt} \right)^3},$$ (11.6)

and so on for higher derivatives. This transformation is called changing the independent variable from $x$ to $t$. It is usually better to work out examples by the methods illustrated above rather than by using the formulas deduced.

Example 11.3.1   Change the independent variable from $x$ to $t$ in the differential equation

$$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0$$

where $x = e^t$.

Solution. $\frac{dx}{dt}= e^t$, therefore

$$\frac{dt}{dx} = e^{-t}\ .$$

Also $\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx}$; therefore $\frac{dy}{dx} = e^{-t} \frac{dy}{dt}$. Also $\frac{d^2 y}{dx^2} = e^{-t} \frac{d}{dx} \left( \frac{dy}{dt} \right) - \fra... ...left( \frac{dy}{dt} \right) \frac{dt}{dx} - \frac{dy}{dt} e^{-t} \frac{dt}{dx}$. Substituting into the last result $\frac{dt}{dx}= e^{-t}$,

$$\frac{d^2 y}{dx^2} = e^{-2t} \frac{d^2 y}{dt^2} - \frac{dy}{dt} e^{-2t}.$$

Substituting these into the differential equation,

$$e^{2t} \left( e^{-2t} \frac{d^2 y}{dt^2} - \frac{dy}{dt} e^{-2t} \right) + e^t \left( e^{-t} \frac{dy}{dt} \right) + y = 0,$$

and reducing, we get $\frac{d^2 y}{dt^2} + y = 0$.

Since the formulas deduced in the Differential Calculus generally involve derivatives of y with respect to $x$, such formulas as the chain rule are especially useful when the parametric equations of a curve are given. Such examples were given in §6.5, and many others will be employed in what follows.