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Old exam questionby skook
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#1
Nov805, 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 


#2
Nov805, 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? 


#3
Nov905, 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...............



#4
Nov905, 08:09 AM

P: 78

Old exam question
I think so, too



#5
Nov905, 04:26 PM

Math
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PF Gold
P: 39,552

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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