# Change of variables of differential equations, and in particular, initial value problems

Transform the given initial value problem into an equivalent problem with the initial point at the origin: $\frac{dy}{dt} = 4 - y^3$, $y(-1) = 2$

I have a feeling that there is an elementary operation that I have failed to learn during my education, seeing as the text writes:

...if some other initial point is given, then we can always make a preliminary change of variables, corresponding to a translation of the coordinate axes, that will take the given point $(t_0, y_0)$ into the origin.

I'd appreciate it if anyone could spoonfeed me on this point, and in particular, describe how one transforms the given problem to

$\frac{dw}{ds} = 4 - (w + 2)^3$, $w(0) = 0$

The condition $$y(-1)=2$$ means that the solution to your ODE passes through the point $$(-1,2)$$ on the $$(t,y)$$ plane.

You want to "move" your axis so that the initial point passes through the point $$(0,0)$$ instead. In order to do this, the $$t$$ axis has to "move 1 unit to the right" and the $$y$$ axis has to move "2 units down". Can you visualize this translation?

The translation is then given by the following set of equations $$\tag{1} \left\{ {\begin{array}{*{20}{c}}{s = t + 1}\\{\omega = y - 2}\end{array}} \right.$$

So now, on the "new" $$(s,\omega)$$ plane your initial point is $$(0,0)$$. You can verify this by plugging $$t=-1$$ and $$y=2$$ in $$(1)$$.

Now, you want to adjust the differential equation in respect to the new variables, that is you want to find the expression for $$\frac{d \omega}{ds}$$ in terms of $$\frac{dy}{dt}$$ and substitute $$t$$ and $$y$$ in terms of $$s$$ and $$\omega$$, respectively. To do this you resort to the chain rule. I'm going to do it in several steps, so it's easier to understand.

$$\tag{2} \frac{{d\omega }}{{dt}} = \underbrace {\frac{{d\omega }}{{dy}}}_{ = 1}\frac{{dy}}{{dt}} = \frac{{dy}}{{dt}}$$

$$\tag{3} \frac{{d\omega }}{{ds}} = \underbrace {\frac{{d\omega }}{{dt}}}_{\frac{{dy}}{{dt}}}\underbrace {\frac{{dt}}{{ds}}}_{ = 1} = \frac{{dy}}{{dt}}$$

In $$(3)$$ I've used the fact that $$s = t + 1 \Leftrightarrow t = s - 1$$ to calculate $$\frac{dt}{ds}$$.

Noting that $$\omega = y - 2 \Leftrightarrow y = \omega + 2$$, we arrive at $$\frac{{d\omega }}{{ds}} = \frac{{dy}}{{dt}} = 4 - {y^3} = 4 - {\left( {\omega + 2} \right)^3}.$$

And that's it! Through this change of variables we have transformed the inital ODE problem into: $$\left\{ \begin{array}{l}\frac{{d\omega }}{{ds}} = 4 - {y^3} = 4 - {\left( {\omega + 2} \right)^3}\\\omega \left( 0 \right) = 0\end{array} \right. .$$

• Nice and clear solution. In the first sentence though, I believe you mean (t, y) plane not (t, x) plane May 22, 2020 at 6:49
• Could someone please explain the use of the chain rule a little more simply to me? I understand that you have to "find the expression for $\frac{d \omega}{ds}$ in terms of $\frac{dy}{dt}$" but I don't understand how that translates into eqs (2) and (3). I don't understand the reasoning behind where these equations come from or what they're meant to accomplish. Thanks!
– Mark
Jul 3, 2021 at 16:42