Old exam question

skook

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

Benny

[tex]
\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
[/tex]

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

skook

I hope the solution is [tex]y=-\frac{x}{\ln{Cx}}[/tex]. 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...............

armandowww

I think so, too

HallsofIvy

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

Then y= vx so [itex]\frac{dy}{dx}= x\frac{dv}{dx}+ v[/itex]

Your equation becomes
[tex]x\frac{dv}{dx}- v= v^2[/tex] or
[tex]x\frac{dv}{dx}= v^2+ v[/tex]
a separable equation.