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Old exam question

by skook
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skook
#1
Nov8-05, 07:08 AM
P: 15
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
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Benny
#2
Nov8-05, 07:29 AM
P: 585
[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
#3
Nov9-05, 07:09 AM
P: 15
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
#4
Nov9-05, 08:09 AM
P: 78
Old exam question

I think so, too
HallsofIvy
#5
Nov9-05, 04:26 PM
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PF Gold
P: 39,552
Quote Quote by 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
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.


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