Sarahisme Posted May 10, 2006 Posted May 10, 2006 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
Sarahisme Posted May 10, 2006 Author Posted May 10, 2006 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?
Sarahisme Posted May 11, 2006 Author Posted May 11, 2006 dont worry think i got it, exact equations right? yay!
Sarahisme Posted May 11, 2006 Author Posted May 11, 2006 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
Sarahisme Posted May 11, 2006 Author Posted May 11, 2006 when they say sketch solution curves do you think they mean draw a phase portrait or something else?
Severian Posted May 11, 2006 Posted May 11, 2006 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. )
Sarahisme Posted May 12, 2006 Author Posted May 12, 2006 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]
Sarahisme Posted May 13, 2006 Author Posted May 13, 2006 so i still think its right.....but i guess it isnt hey?
Severian Posted May 13, 2006 Posted May 13, 2006 Sorry, I think you are right with that solution. My mistake.
Sarahisme Posted May 15, 2006 Author Posted May 15, 2006 thanks for all the btw Severian. much appreciated.
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