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anonymous
 one year ago
The slope field for a differential equation is shown in the figure. Determine the general solution of this equation
anonymous
 one year ago
The slope field for a differential equation is shown in the figure. Determine the general solution of this equation

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0answer choices y=Cx^2 x=Cy^2 x^2 – y^2 = C^2 x2 + y2 = C^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so i know i have to reverse engineer this but i'm not doing too well

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439493039926:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i should just isolate the variables of each of the answer choices and derive each side right and graph all the diff equations i get and match it to the one im given , is that right?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Could you not let x = 1 and y = 1 and then set it up as dy/dx that way you can see the slope easier?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4To my knowledge that slope field corresponds to a differential equation : \[y'=2x\] the family of solutions must look something like \[y = x^2+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.mathscoop.com/calculus/differentialequations/slopefieldgenerator.php this might help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and yes you are correct @ganeshie8 i just cant get the differential equation when onlly given the solution to a differential equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay can you walk me through your answer ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Wait, I don't know if it's right, but what I was thinking is as I said above, let x = 1, y = 1 or what ever and just let dy/dx = x/y^2 and then plug in the values and just observe the points...not sure if that's a good method haha, ganeshie definitely knows more.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the solution curves are the smooth curves obtained by joining the slope segments in the slope field : dw:1439494028215:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Easy to see that they are family of parabolas that open up, under vertical translation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay let me rephrase the question, if i give you the solution to a differential equation (say x=Cy^2) can you find the differential equation it came from ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol i know i dont know how to make the constant multiply :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4x = Cy^2 1 = C2yy' x/1 = y/(2y') y' = y/(2x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay and if we graph that we get

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so b is not our answer, so know tell me how you did that so i can do it to the rest of the answer choices XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4since you have "one" arbitrary constant, you want to differentiate the solution just once

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so from x = Cy^2 to 1 = C2yy' the only thing you did was derive each side right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4right, just implicitly differentiate both sides with respect to x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what did you do from 1 = C2yy' to x/1 = y/(2y') ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4x = Cy^2 (1) 1 = C2yy' (2) next divide : (1)/(2)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4your goal is to get rid of that arbitrary constant C dividing both equations does just that

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Nice explanation @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i see so what do you do after you vet rid of the c ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you're done! the equation w/o arbitrary constant is the mother "differential equation" that represents the "family of solution curves" : x = Cy^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so let me try to do that for a different answer choice and tell me if i'm way off.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the differential equation y' = y/(2x) represents the whole family of curves x = Cy^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Btw, none of the options are correct, but if I were you, I would tick option A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um okay what makes you chose option 1 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4The correct answer is \(y = x^2 + C\) From the slope field it is clear that we must get solution curves that look same upto a vertical translation. My guess is that your professor cooked up this problem carelessly... Option 1 is close to parabolas, so..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay sounds about right :P , will let you know if its right in a little while when i turn it in

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4good luck! please do share what the grader thinks is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0will do ! thanks for all your help guys ! :D

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you can rewrite these things as level surfaces and i think you can nail it pretty solidly so if you take option B that has been discounted, you can say: \(x=Cy^2 \implies f(x,y) = x  Cy^2 = constant\) \(grad f = <1, 2Cy>\) so the slope \(dy/dx = \frac{1}{2Cy} = \frac{1}{2Cy}\) and it can't be that because dy/dx has no dependency on y and doesn't go ballistic at y = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay are you suggesting its option 1 then ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i'd go A i can justify it, just lost a load of latex, so i might try post it again later

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1here's the observation: if you write \(y = mx + c\) as a closed surface , you get \(f(x,y) = y  mx = const\) so \(grad(f) = <m, 1>\) so slope in normal xyspeak is \(\frac{f_x}{f_y}\) that's not that surprising as \(df = f_x dx + f_y dy = 0 \implies \frac{dy}{dx} = \frac{f_x}{f_y}\) do that for \(y=Cx^2\) and i think you get something that behaves like your graph.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was right !! :D thanks again !!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i reckon the "actual" right answer is: \(y=\frac{x^2}{2}+C \)
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