A community for students.
Here's the question you clicked on:
 0 viewing
huclogin
 one year ago
The problem is that find the curve, its slope of tangent lines are twice as steep as the ray from the origin. The point is (x,y)
I solve it like this:(dy/dx)=(2y/x) ∫dy/y=∫dx/x
then lny=2lnx+c (c is constant) then y=Ax^2(A=±e^c)
I think that A is ambiguous up to the constant c, and the graph of the curve I'm aiming for, and the graph of the curves is a bunch of parabolas which is undefined at point (0,0). Did I make any mistakes? If I have any one of the curves to the left of yaxis and to the right connected together and take the derivative of it, can I always get dy/dx=2y/x?
huclogin
 one year ago
The problem is that find the curve, its slope of tangent lines are twice as steep as the ray from the origin. The point is (x,y) I solve it like this:(dy/dx)=(2y/x) ∫dy/y=∫dx/x then lny=2lnx+c (c is constant) then y=Ax^2(A=±e^c) I think that A is ambiguous up to the constant c, and the graph of the curve I'm aiming for, and the graph of the curves is a bunch of parabolas which is undefined at point (0,0). Did I make any mistakes? If I have any one of the curves to the left of yaxis and to the right connected together and take the derivative of it, can I always get dy/dx=2y/x?

This Question is Closed

phi
 one year ago
Best ResponseYou've already chosen the best response.0it sounds right. though the parabola is defined at (0,0) however, we may want to exclude the origin?

phi
 one year ago
Best ResponseYou've already chosen the best response.0because the "ray from the origin" is undefined at (0,0)

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438353168811:dw It's the graph. I can not draw very well.

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1Oh.. Maybe I can't call it parabolas, it's only similar to parabolas. Because when I solve the differential equations I divided it by x on the righthand side.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2The domain must be restricted to either x<0 or x>0 because the slope is undefined at x=0. From the slope field you can see that as you travel along a particular curve toward 0, the moment you reach 0, you wont know which direction to go because there are too many directions : dw:1438353913296:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.0ganeshie8 are you suggesting the solution is Ax^2 x<0 Bx^2 x>0 where A and B are nonzero constants.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Nope, domain depends on the initial value If the initial value starts negative, we must restrict the domain to x<0 because we wont know which direction that particular solution goes after that singularity at x=0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2solution curves are half parabolas

phi
 one year ago
Best ResponseYou've already chosen the best response.0that sounds like philosophy. you could glue two different parabolas together, make sure one of them goes through the initial value, exclude (0,0) and you have a "solution" that meets the conditions.

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1ganeshie8 Do you meas after I take the indefinite integral of the differential equation dy/y=2dx/x I can get a family of functions?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, the solutions are family of half parabolas, not full parabolas with a hole at (0, 0)

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1ganeshie8 dw:1438354495749:dw If I connect the two curves(any curves, one from the left and one from the right) to gether and get a function then take the derivative of it, can I always get dy/dx=2y/x? Another question, I think that A is ambiguous up to the constant c, is it right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes for both, \(A=e^C\) is as much an arbitrary constant as \(C\) is

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2but wait, do we need to worry that A is now positive always ?

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1ganeshie8 NO, because after taking the integral of dy/y we get lny

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ahh right, that does the trick : \[\ln y = \lnx^2 + C \implies y = e^Cx^2 \implies y = \pm e^Cx^2 = Ax^2 \] \(A\) can be negative too !

huclogin
 one year ago
Best ResponseYou've already chosen the best response.1ganeshie8 phi My problem is solved now.Thank you! ^ ^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.