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Given continuous $f: \mathbb{R} \rightarrow \mathbb{R}$, $\begin{vmatrix}a_1 & b_1 \\ a_2 & b_2\end{vmatrix} \neq 0$ and consider the ODE

$$y'(t) = f\left(\frac{a_1 t + b_1 y + c_1}{a_2 t + b_2 y + c_2}\right)$$

I'm asked to solve the given ODE by transforming it to an homogenous ODE (in german: auf ein homogenes System zurückführen). This means that there is a function $h: \mathbb{R} \rightarrow \mathbb{R}$ such that the equation above is equivalent to

$$y'(t) = h\left(\frac{y(t)}{t}\right)$$

Currently I'm stuck and don't have a clue how to approach this. It would be enough to show that one can express the fraction in the initial equation as a function of only $\frac{y(t)}{t}$, but I can't think of a good approach and don't know how to factor in the non-zero determinant.

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  • $\begingroup$ I have taken the liberty to modify your title in order for it to be easily understood (for example, the fact that the determinant is nonzero is superfluous in the title) $\endgroup$
    – Jean Marie
    Commented Mar 18, 2020 at 18:27
  • $\begingroup$ @JeanMarie Thank you. This title can be easily understood indeed. $\endgroup$ Commented Mar 18, 2020 at 18:48
  • $\begingroup$ One notational issue: the second equation should read $y'(t) = h(y / t)$, omitting the dependence of $y$ on $t$. The point is that the ODE reads $y'(t) = g(y(t), t)$ where $g(y, t)$ a function of $y/t$. It's misleading to think of $g(y, t)$ as a function of $y(t) / t$; this suggests that we should calculate, for each $t$, a value of $y$ so that $g(y(t), t) = h(y(t) / t)$, which is what the other answer to this question has done. This latter approach seems wrong to me. $\endgroup$ Commented Mar 18, 2020 at 20:32
  • $\begingroup$ @ChristopherGadzinski Sorry. I'm not exactly sure if I understand your point correctly. I though about it like this: $h$ should be a function with domain $\mathbb{R}$ with input $y(t)/t$ (or $y/t$ - and don't see why this makes a difference), in comparison to a function $f$ which takes two inputs. $\endgroup$ Commented Mar 18, 2020 at 20:46
  • $\begingroup$ The issue is subtle, but there are two things that I don't like: 1) In the rhs of the first equation you're writing $y$ without the $t$-dependence explicit, which might suggest that the role of $y$ in rhs of the second equation is different from its role in the first. 2) In your last paragraph, one should talk about writing the rational function in the first equation "as a function of only $y/t$" rather than of $y(t)/t$; the latter is itself a function of $t$! $\endgroup$ Commented Mar 18, 2020 at 21:02

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This equation would be homogeneous if $c_1 = c_2 = 0$. However, it is not quite homogeneous in general. What we can do is translate the variables, setting $\tilde t = t - t_0$ and $\tilde y = y - y_0$, so that $$\begin{cases}a_1 \tilde t + b_1 \tilde y = a_1 t + b_1 y + c_1 \\ a_2 \tilde t + b_2 \tilde y = a_2 t + b_2 y + c_2\end{cases}$$ In terms of $t_0$ and $y_0$ this is a system of two linear equations, and the condition on the determinant we are given guarantees its non-singularity. With respect to the translated coordinates, we get $$\frac{d}{d \tilde t} \tilde y = \frac{d}{dt} y = f\left(\frac{a_1 \tilde t + b_1 \tilde y}{a_2 \tilde t + b_2 \tilde y}\right)$$ which is homogeneous.

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  • $\begingroup$ I think this is, what the solution is intended to look like. The exact formulation (from German) is "Look at the ODE [...] and solve it via a homogenous Equation." In particular the statement is not exactly. "Show that ODE [...] is homogeneous.". My question was kind of misleading in the formulation here. $\endgroup$ Commented Mar 18, 2020 at 20:17
  • $\begingroup$ To summarise your approach: 1. We find a linear transformation of $(y, t)$ such that the equation is homogenous using the new coordinates. 2. Solve the now homogenous solution. 3. Use the inverse transformation to obtain the solution of the given ODE. Is this correct? $\endgroup$ Commented Mar 18, 2020 at 20:19
  • $\begingroup$ Yes, that's the idea. $\endgroup$ Commented Mar 18, 2020 at 20:35
  • $\begingroup$ Differentiation with respect to $\tilde t$ is exactly the same as differentiation with respect to $t$! In one dimension, we have the law $\frac{d}{dx} = \frac{d}{dy} \frac{dy}{dx}$. $\endgroup$ Commented Mar 18, 2020 at 20:51
  • $\begingroup$ (In general, we can write $\frac{d}{dx} = \sum_i \frac{d}{d y_i} \frac{d y_i}{dx}$ for any vector field $\frac{d}{dx}$ and coordinate chart $y_i$.) $\endgroup$ Commented Mar 18, 2020 at 21:08

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