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critical points...


Sarahisme

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hey, sorry for all the questions lately :P but ....

 

for this is question, is the only critical point (1,0) ? it keeps refering to critical pointS in the problem... but i think there is the only the one critical point, right?

 

as for the rest of the question, well i suppose i need to figure this little kenumdrem (spelling? :P) out first, hey :P

 

picture144ml.png

 

thanks, and apologies for the questions again! :cool:

 

-Sarah

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hmm got (a)

 

its part (d) that i can't seem to get now.... how can you integrate this:

 

[math] \frac{dy}{dx} = \frac{1-x^2+y^2}{2xy} [/math]

 

i am assuming thats how you find the 'complete solution' right? :P

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hmm well actually i got the equation

 

[math] x + \frac{y^2+1}{x} = c [/math]

 

which produces a contour plot like this

 

mmmmmm4en.jpg

 

but that doesnt seem to relate to the critical points found hmmm..... me confuseled :P

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Describe to us how you got your 'complete solution'. (Hint: your equation [math] x + \frac{y^2+1}{x} = c [/math] is not a solution of [math] \frac{dy}{dx} = \frac{1-x^2+y^2}{2xy} [/math]. To see why, solve our solution for y and differentiate. )

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ok i'll explain how i got it....

 

i used exact equations method. i.e. [math] M_y = N_x [/math]

 

rearranging the equation to :

 

[math] (x^2-y^2-1) + (2xy)y' = 0 [/math]

 

but the equation is not exact to start with so have to find integrating factor.

 

i found this integrating factor to be [math] \frac{1}{x^2} [/math]

 

multiplying this through the original equation gives

 

[math] (1-\frac{y^2}{x^2}-\frac{1}{x^2}) + (\frac{2y}{x})y' = 0 [/math]

 

the equation is now exact.

 

and so solving it gives the answer i found:

 

[math]

x + \frac{y^2+1}{x} = c

[/math]

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