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

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

Mathematics
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answer choices y=Cx^2 x=Cy^2 x^2 – y^2 = C^2 x2 + y2 = C^2
okay so i know i have to reverse engineer this but i'm not doing too well

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|dw:1439493039926:dw|
so 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?
Could you not let x = 1 and y = 1 and then set it up as dy/dx that way you can see the slope easier?
To my knowledge that slope field corresponds to a differential equation : \[y'=2x\] the family of solutions must look something like \[y = x^2+C\]
http://www.mathscoop.com/calculus/differential-equations/slope-field-generator.php this might help
and yes you are correct @ganeshie8 i just cant get the differential equation when onlly given the solution to a differential equation
okay can you walk me through your answer ?
Wait, 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.
the solution curves are the smooth curves obtained by joining the slope segments in the slope field : |dw:1439494028215:dw|
Easy to see that they are family of parabolas that open up, under vertical translation
okay 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 ?
lol i know i dont know how to make the constant multiply :P
x = Cy^2 1 = C2yy' x/1 = y/(2y') y' = y/(2x)
okay and if we graph that we get
so 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
since you have "one" arbitrary constant, you want to differentiate the solution just once
okay so from x = Cy^2 to 1 = C2yy' the only thing you did was derive each side right ?
right, just implicitly differentiate both sides with respect to x
so what did you do from 1 = C2yy' to x/1 = y/(2y') ?
x = Cy^2 ---------------(1) 1 = C2yy' --------------(2) next divide : (1)/(2)
okay umm why ?
your goal is to get rid of that arbitrary constant C dividing both equations does just that
Nice explanation @ganeshie8
okay i see so what do you do after you vet rid of the c ?
get*
you're done! the equation w/o arbitrary constant is the mother "differential equation" that represents the "family of solution curves" : x = Cy^2
okay so let me try to do that for a different answer choice and tell me if i'm way off.
the differential equation y' = y/(2x) represents the whole family of curves x = Cy^2
Btw, none of the options are correct, but if I were you, I would tick option A
um okay what makes you chose option 1 ?
The 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..
okay sounds about right :P , will let you know if its right in a little while when i turn it in
good luck! please do share what the grader thinks is right
will do ! thanks for all your help guys ! :D
you can re-write 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
okay are you suggesting its option 1 then ?
i'd go A i can justify it, just lost a load of latex, so i might try post it again later
here'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 xy-speak 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.
i was right !! :D thanks again !!
i reckon the "actual" right answer is: \(y=\frac{x^2}{2}+C \)
i mean it not i

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