# Old exam question

by skook
Tags: exam
 P: 15 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
 P: 585 $$\frac{{dy}}{{dx}} - \frac{y}{x} = \frac{{y^2 }}{{x^2 }} \Rightarrow \frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right) + \left( {\frac{y}{x}} \right)^2$$ Let y = v(x)x. Is this a standard substitution for the subject you are studying?
 P: 15 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). First maths course I've done in over 25 years...............
P: 78

## Old exam question

I think so, too
Math
Emeritus
Thanks
PF Gold
P: 38,879
 Quote by 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
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
$$x\frac{dv}{dx}= v^2+ v$$
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