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Consider the differential equation...

$y' = f(\frac{y}{t})$

Show that the substitution $v = \frac{v}{t}$ leads to a separable differential equation in $v$

Here's what I did.

$v = \frac{y}{t}$

$\frac{dv}{dt} = \frac{dy}{dt} - \frac{1}{t^2}$

Sub into the orignal.

$\frac{dv}{dt} = f(v) - \frac{1}{t^2}$

$\frac{dv}{f(v)} =\frac{1}{t^2}dt$

This is where I get stuck. Is this the what the question is asking for, or am I forgetting to do something?

1 Answers 1

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Note that if you substitute $v=\frac{y}{t} \Rightarrow y=vt$. Note that $v$ is assumed to be a function of $t$. Using the product rule, we obtain the derivative $y'$: $$y=v(t)\cdot t \Rightarrow y'=\frac{dv}{dt}\cdot t+v \tag{1}$$ Substituting into $y'=f\left(\frac{y}{t}\right)$: $$\frac{dv}{dt}\cdot t+v=f(v)$$ Now, try to separate the variables. i.e, put only $v$ and $dv$ terms on one side of the equality and $t$ and $dt$ terms on the other side, then you are done.

Feel free to ask on the comments if you have any related doubts or questions.

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    is $\frac{t}{dt}= \frac{f(v) - v}{dv}$ okay? Are $dt$ and $dv$ allowed to be on the bottom of the fraction? Also, can $f(v)$ be expressed in another way? Like $v$ alone, or no? – 2017-02-13
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    You are one step away! Just take the reciprocal on both sides, then you are done. No, $f(v)$ should be left like this. – 2017-02-13
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    $\frac{dt}{t}= \frac{dv}{f(v) - v}$ ? – 2017-02-13
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    **Yes, that is correct.** Therefore, your differential equation is separable. By the way, to answer your question about $f(v)$: $f(v)$ should be left as it is. Otherwise, it loses its generality. For example if you say $f(v)$, it can mean **any function of $v$**. If you just say $v$, it means $v$ **only**. – 2017-02-13