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Here is a statement (or theorem) from my book:

$\pmb{\text{A neat trick: Turning nonlinear separable equations into linear separable equations}}$

$\mathcal{\text{Knowing when to substitute:}}$

You can use the trick of setting $y=xv$ when you have differential equation that is of the form:

$\dfrac{dy}{dx}=f(x,y)$

when $f(x,y)=f(tx,ty)$ where $t$ is a constant.

I understand this. But the book gives no proof. I mean how can we ensure that the trick of setting $y=xv$ will always work whenever $f(x,y)=f(tx,ty)$

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  • $\begingroup$ What have you tried ? $\endgroup$ – Yves Daoust Feb 2 at 13:58
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Just plug $y=xv$ in the equation.

$$\frac{d(xv)}{dx}=x\frac{dv}{dx}+v=f(x,xv)$$

and

$$\frac{dv}{f(1,v)-v}=\frac{dx}x.$$


Alternatively,

$$f(tx,ty)=f(x,y)\iff f(x,y)=g\left(\frac yx\right)=g(v)$$ and the equation is

$$xv'+v=g(v).$$

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  • $\begingroup$ If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $\dfrac{y}{x}$? $\endgroup$ – Oliver Feb 2 at 15:32
  • $\begingroup$ @Oliver: hem, $t=1/x$. $\endgroup$ – Yves Daoust Feb 2 at 16:10
  • $\begingroup$ Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant. $\endgroup$ – Oliver Feb 2 at 16:21
  • $\begingroup$ @Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)\to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant... $\endgroup$ – Yves Daoust Feb 2 at 16:26

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