# Old exam question

1. Nov 8, 2005

### skook

Could someone please just give me a hint to get started.
$$\frac{dy}{dx}-\frac{y}{x}=\frac{y^2}{x^2} for x>0$$
thanks
Skook

2. Nov 8, 2005

### Benny

$$\frac{{dy}}{{dx}} - \frac{y}{x} = \frac{{y^2 }}{{x^2 }} \Rightarrow \frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right) + \left( {\frac{y}{x}} \right)^2$$

Let y = v(x)x. Is this a standard substitution for the subject you are studying?

3. Nov 9, 2005

### skook

thanks for that

I hope the solution is $$y=-\frac{x}{\ln{Cx}}$$. It was from an Open University course (http://www3.open.ac.uk/courses/bin/p12.dll?C02MS324) [Broken]. First maths course I've done in over 25 years...............

Last edited by a moderator: May 2, 2017
4. Nov 9, 2005

### armandowww

I think so, too

5. Nov 9, 2005

### HallsofIvy

Staff Emeritus
Since x and y only appear together as y/x, try the obvious substitution: Introduce a new dependent variable $v= \frac{y}{x}$.

Then y= vx so $\frac{dy}{dx}= x\frac{dv}{dx}+ v$

$$x\frac{dv}{dx}- v= v^2$$ or
$$x\frac{dv}{dx}= v^2+ v$$