Interchange of dependent and independent variables

It is sometimes desirable to transform an expression involving derivatives of $y$ with respect to $x$ into an equivalent expression involving instead derivatives of $x$ with respect to $y$. Our examples will show that in many cases such a change transforms the given expression into a much simpler one. Or perhaps $x$ is given as an explicit function of $y$ in a problem, and it is found more convenient to use a formula involving $\frac{dx}{dy}$, $\frac{d^2 x}{dy^2}$, etc., than one involving $\frac{dy}{dx}$, $\frac{d^2 y}{dx^2}$, etc. We shall now proceed to find the formulas necessary for making such transformations.

Given $y = f(x)$, then from item XXVI in §5.1, we have

$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}, \ \ \ \frac{dx}{dy} \ne 0$$ (11.1)

giving $\frac{dy}{dx}$ in terms of $\frac{dx}{dy}$. Also, by XXV in §5.1,

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

or

$$\frac{d^2 y}{dx^2} = \frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) \frac{dy}{dx}.$$ (11.2)

But $\frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) = - \frac{\frac{d^2 x}{dy^2}}{\left( \frac{dx}{dy} \right)^2}$; and $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ from (11.1). Substituting these in (11.2), we get

$$\frac{d^2 y}{dx^2} = -\frac{ \frac{d^2 x}{dy^2} }{\left( \frac{dx}{dy} \right)^3},$$ (11.3)

giving $\frac{d^2 y}{dx^2}$ in terms of $\frac{dx}{dy}$ and $\frac{d^2 x}{dy^2}$. Similarly,

$$\frac{d^3 y}{dx^3} = -\frac{ \frac{d^3 x}{dy^3} \frac{dx}{dy} - 3 \left( \frac{d^2 x}{dy^2} \right)^2 }{ \left( \frac{dx}{dy} \right)^5 },$$ (11.4)

and so on for higher derivatives. This transformation is called changing the independent variable from $x$ to $y$.

Example 11.1.1   Change the independent variable from $x$ to $y$ in the equation

$$3 \left( \frac{d^2 y}{dx^2} \right)^2 - \frac{dy}{dx} \frac{d^3 y}{dx^3} - \frac{d^2 y}{dx^2} \left( \frac{dy}{dx} \right)^2 = 0.$$

Solution. Substituting from (11.1), (11.3), (11.4),

$$3 \left( -\frac{ \frac{d^2 x}{dy^2} }{ \left( \frac{dx}{dy} \rig... ...c{dx}{dy} \right)^3 } \right) \left( \frac{1}{ \frac{dx}{dy} } \right)^2 = 0.$$

Reducing, we get

$$\frac{d^3 x}{dy^3} + \frac{d^2 x}{dy^2} = 0,$$

a much simpler equation.