# Elementary differential equations: Please give a proof of the trick

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 $$f(x,y)=f(tx,ty)$$

• What have you tried ? – Yves Daoust Feb 2 at 13:58

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

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

• If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $\dfrac{y}{x}$? – Oliver Feb 2 at 15:32
• @Oliver: hem, $t=1/x$. – Yves Daoust Feb 2 at 16:10
• 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. – Oliver Feb 2 at 16:21
• @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... – Yves Daoust Feb 2 at 16:26