1. Constant solution

I want to find constant solutions of the differential equation $\frac{dy}{dx}=\frac{y^3}{x^2}-2\frac{y}{x}$. I set $y=c$, where $c$ is a constant. Then $\frac{dy}{dx}=0$ and the differential equation becomes

$0=\frac{c^3}{x^2}-2\frac{c}{x}$
$\frac{c}{x}(\frac{c^2}{x}-2)=0$
$c/x=0$ or $c^2/x-2=0$
$c=0$ or $c^2=2x$

How to explain in words the equation $c^2=2x$ does not give any constant solution of the differential equation?

2. Re: Constant solution

Originally Posted by woo
I want to find constant solutions of the differential equation $\frac{dy}{dx}=\frac{y^3}{x^2}-2\frac{y}{x}$. I set $y=c$, where $c$ is a constant. Then $\frac{dy}{dx}=0$ and the differential equation becomes

$0=\frac{c^3}{x^2}-2\frac{c}{x}$
$\frac{c}{x}(\frac{c^2}{x}-2)=0$
$c/x=0$ or $c^2/x-2=0$
$c=0$ or $c^2=2x$

How to explain in words the equation $c^2=2x$ does not give any constant solution of the differential equation?
$c^2 = 2x \rightarrow x = \frac{c^2}{2}$

Looks like a constant solution to me.

-Dan

3. Re: Constant solution

No it's not. It's a solution that works for only one value of $x$!

I would say that the left hand side of $c^2=2x$ is constant and has only one value while the right hand side is variable and can take many values.