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

1. anonymous

2. anonymous

answer choices y=Cx^2 x=Cy^2 x^2 – y^2 = C^2 x2 + y2 = C^2

3. anonymous

okay so i know i have to reverse engineer this but i'm not doing too well

4. ganeshie8

|dw:1439493039926:dw|

5. anonymous

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?

6. Astrophysics

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?

7. ganeshie8

To my knowledge that slope field corresponds to a differential equation : $y'=2x$ the family of solutions must look something like $y = x^2+C$

8. anonymous
9. anonymous

and yes you are correct @ganeshie8 i just cant get the differential equation when onlly given the solution to a differential equation

10. anonymous

11. Astrophysics

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.

12. ganeshie8

the solution curves are the smooth curves obtained by joining the slope segments in the slope field : |dw:1439494028215:dw|

13. ganeshie8

Easy to see that they are family of parabolas that open up, under vertical translation

14. anonymous

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 ?

15. anonymous

lol i know i dont know how to make the constant multiply :P

16. ganeshie8

x = Cy^2 1 = C2yy' x/1 = y/(2y') y' = y/(2x)

17. anonymous

okay and if we graph that we get

18. anonymous

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

19. ganeshie8

since you have "one" arbitrary constant, you want to differentiate the solution just once

20. anonymous

okay so from x = Cy^2 to 1 = C2yy' the only thing you did was derive each side right ?

21. ganeshie8

right, just implicitly differentiate both sides with respect to x

22. anonymous

so what did you do from 1 = C2yy' to x/1 = y/(2y') ?

23. ganeshie8

x = Cy^2 ---------------(1) 1 = C2yy' --------------(2) next divide : (1)/(2)

24. anonymous

okay umm why ?

25. ganeshie8

your goal is to get rid of that arbitrary constant C dividing both equations does just that

26. Astrophysics

Nice explanation @ganeshie8

27. anonymous

okay i see so what do you do after you vet rid of the c ?

28. anonymous

get*

29. ganeshie8

you're done! the equation w/o arbitrary constant is the mother "differential equation" that represents the "family of solution curves" : x = Cy^2

30. anonymous

okay so let me try to do that for a different answer choice and tell me if i'm way off.

31. ganeshie8

the differential equation y' = y/(2x) represents the whole family of curves x = Cy^2

32. ganeshie8

Btw, none of the options are correct, but if I were you, I would tick option A

33. anonymous

um okay what makes you chose option 1 ?

34. ganeshie8

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..

35. anonymous

okay sounds about right :P , will let you know if its right in a little while when i turn it in

36. ganeshie8

37. anonymous

will do ! thanks for all your help guys ! :D

38. IrishBoy123

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

39. anonymous

okay are you suggesting its option 1 then ?

40. IrishBoy123

i'd go A i can justify it, just lost a load of latex, so i might try post it again later

41. IrishBoy123

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.

42. anonymous

i was right !! :D thanks again !!

43. IrishBoy123

i reckon the "actual" right answer is: $$y=\frac{x^2}{2}+C$$

44. anonymous

i mean it not i