hey, sorry for all the questions lately 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? ) out first, hey

 

 

thanks, and apologies for the questions again!

 

-Sarah

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?

dont worry think i got it, exact equations right? yay!

hmm well actually i got the equation

 

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

 

which produces a contour plot like this

 

 

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

when they say sketch solution curves do you think they mean draw a phase portrait or something else?

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. )

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]

so i still think its right.....but i guess it isnt hey?

Sorry, I think you are right with that solution. My mistake.

ok.

thanks for all the btw Severian. much appreciated.