Dependent variable substitution of a differential equation.

Question

I am attempting to answer a question from a textbook. The question is as follows:

"Use the substitution $y = x^2$ to turn the differential equation $x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$ into a second order differential equation with constant coefficients involving y and t."

The answer according to the textbook is $\frac{d^2y}{(dt)^2} + \frac{dy}{dt} = 0$

I am unsure what to do with the $(\frac{dx}{dt})^2$ term in the original equation. Is it equivalent to $\frac{d}{dt}(x^2)$? If this is the case then my working gets as far as the following before I get stuck:

$x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$

$\Rightarrow x\frac{d^2x}{(dt)^2} + \frac{d}{dt}(x^2) + x\frac{dx}{dt} = 0$

$\frac{dy}{dt} =\frac{dy}{dx}\frac{dx}{dt}=2x\frac{dx}{dt} \Rightarrow \frac{d^2x}{(dt)^2}=\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt})$

$\Rightarrow x\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt}) + 2x + x\frac{1}{2x}\frac{dy}{dt} = 0$

$\Rightarrow \frac{1}{2x}\frac{dy}{dt} + \frac{1}{2}\frac{d^2y}{(dt)^2} + 2x + \frac{1}{2}\frac{dy}{dt} = 0$

Please can someone show me where I have gone wrong?

Answer 1

In general, you are incorrect that $\left(\frac{dx}{dt}\right)^2 = \frac{d}{dt}\left(x^2\right)$ since $\frac{d}{dt}\left(x^2\right) = 2x\left(\frac{dx}{dt}\right)$, so they're equal only in the special case where $2x = \frac{dx}{dt}$. Instead, multiply the original equation on both sides by $2$ to get

$$2x\frac{d^2x}{(dt)^2} + 2\left(\frac{dx}{dt}\right)^2 + 2x\frac{dx}{dt} = 0 \tag{1}\label{eq1A}$$

Next, differentiating $y = x^2$ gives

$$\frac{dy}{dt} = 2x\left(\frac{dx}{dt}\right) \tag{2}\label{eq2A}$$

and then differentiating again, using the product rule this time, gives

$$\frac{d^2y}{(dt)^2} = 2\left(\frac{dx}{dt}\right)^2 + 2x\left(\frac{d^2x}{dt^2}\right) \tag{3}\label{eq3A}$$

As you can see, with the left side of \eqref{eq1A}, we have \eqref{eq3A} matching the first $2$ terms and \eqref{eq2A} matching the final third term. Thus, \eqref{eq1A} can be rewritten as

$$\frac{d^2y}{(dt)^2} + \frac{dy}{dt} = 0 \tag{4}\label{eq4A}$$

Answer 2

As noticed $$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt}\cdot \frac{dx}{dt}\neq \frac{d(x^2)}{dt}$$

By the suggested substitution we have

$$y=x^2 \implies \frac{dy}{dt}=2x \frac{dx}{dt} \implies \frac{d^2y}{dt^2}=2 \left(\frac{dx}{dt}\right)^2+2x \frac{d^2x}{dt^2}$$

that is

• $\frac{dx}{dt}=\frac1{2x}\frac{dy}{dt} $
• $\frac{d^2x}{dt^2}=\frac1{2x}\frac{d^2y}{dt^2}-\frac1{4x^3}\left(\frac{dy}{dt}\right)^2 $

then

$$x\frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + x\frac{dx}{dt}=0$$

$$\frac1{2}\frac{d^2y}{dt^2}\color{red}{-\frac1{4x^2}\left(\frac{dy}{dt}\right)^2 +\frac1{4x^2}\left(\frac{dy}{dt}\right)^2}+ \frac1{2}\frac{dy}{dt}=0$$

$$\frac1{2}\frac{d^2y}{dt^2}+ \frac1{2}\frac{dy}{dt}=0$$

Answer 3

$$x\frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + x\frac{dx}{dt} = 0$$ With another notation: $$\color {blue}{xx''+ (x')^2} + xx' = 0$$ $$\color {blue}{(xx')'}+xx'=0$$ Multiply by $2$: $$(2xx')'+2xx'=0$$ Since $(x^2)'=2xx'$: $$(x^2)''+(x^2)'=0$$ And as $y=x^2$ this is simply: $$y''+y'=0$$