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Differential equations, orthogonal trajectories

  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data
    you are given a family of curves, in this case i was given a bunch of circles x^2+y^2=cx, sketch these curves for c=0,2,4,6, both positive and negative, solve the equation for c and differentiate both sides with respect to x and solve for dy/dx. You obtain an ODE in the form of dy/dx=f(x,y), now any orthogonal or perpendicular trajectory must be perpendicular at any point, so we're looking for curves satisfying dy/dx=-1/f(x,y) (negative reciprocal slope). Find an implicit solution to this ODE, and sketch these

    2. Relevant equations
    x^2+y^2=cx

    3. The attempt at a solution
    I plotted all the circles, the first one c=0 is a point, and the rest are circles on the left and right sides of the origin, each new circle's radius moves outwards by 1. This was easy part, and solving for c i got c=(x^2+y^2)/x, i split this up into c=x+y^2/x and differentiated to get 0=1-(y^2/x^2)(dy/dx) so that means dy/dx=x^2/y^2. I took the reciprocal so the new ODE becomes dy/dx=-y^2/x^2, but the problem is that when i solve this ODE i get y=Ce^(1/x) and when i plot this for random C values none of these are perpendicular to my circles! I would like to know where my mistake is, did I differentiate in the first step correctly? Where did I go wrong?
     
  2. jcsd
  3. May 12, 2015 #2
    This is the equation if a circle x^2 + y^2 = K,
    So what you hav above is a circle just rearrange to check .
    Differentiation gives 2*x + 2*y*y' = c
    Solve for dy/dx = (c-2x)/(2*y)
    What you need is dy/dx = -(2*y)/(c-2x)
    This is a seperable 1 ODE that can be solve by re-arranging to
    (1/2*y)dy = dx/(c-2x)
    And it's up to you ;) Good luck
     
    Last edited: May 13, 2015
  4. May 12, 2015 #3
    okay i am going to use your work and solve and see if it is perpendicular thanks, but i plotted the original solutions and they came out as circles,i rewrote them as (x-c/2)^2+y^2=(c/2)^2, i will check
     
  5. May 13, 2015 #4
    i guess its the same if you differentiate it without solving for c first
     
  6. May 13, 2015 #5
    I diff the whole equation without re-arrangement, didn't got rid of anything, they are indeed circles,if you take a look, everything is Okay cuz (x^2)' = 2x , (cx)' = c , (y^2)' = 2yy'
     
  7. May 13, 2015 #6
    Yes I understood it, i have now obtained -ln(y)/2 + C = -ln(c-2x)/2 + C i think to plot I have to choose the parameter C, thanks
     
  8. May 13, 2015 #7
    Unless they cancel out, but i dont think they would have the same value because they are different constants
     
  9. May 13, 2015 #8
    wait nvm i got it, but just to make sure can i just get rid of the big constant C's and then all is left is ln(y)=ln(c-2x)
     
  10. May 13, 2015 #9
    Ln(y) = ln(c-2x) + K (K =(C-c)/2 is the integration constants)
    Y = c-2x + Cte (constant of integration)
     
    Last edited: May 13, 2015
  11. May 13, 2015 #10
    why are you subtracting C-c.... and where does division by 2i come from? There arent any irrational numbers..... What the hell is C t and e.. i cant plot this with so many random constants
     
  12. May 13, 2015 #11
    Oops, keyboard typing errors i means C-c/2 = K to treat both of them as one constant C and c are constant if ODE solving
     
  13. May 13, 2015 #12
    Little c is a number though, 0,-2,2,-4,4,etc, im gonna assume you mean C-C/2=K, so i did the same thing to simplify by raising e^ln(), are you sure you can just take an average randomly to get K or it doesnt matter since it doesnt show up, it only shows up as K. So do I pick random K values? This needs to be perpendicular to x^2+y^2=cx
     
  14. May 13, 2015 #13
    Plotting y=c-2x alone gives lines through each circle through origin so that works if i ignore K, thanks for your immense help and clarifying my stupid questions
     
  15. May 13, 2015 #14
    K is an integration constant, in fact i didnt mean to write it as C-c/2 but I can so why not ? it show up as a single constant
     
  16. May 13, 2015 #15
    Welcome, but don't forget to write as y= c-2x + K, in your case K = 0 happened to work, but allways write the most general solution and pick up the values you need, (you picked up K= 0 and that's great )
     
  17. May 13, 2015 #16
    wait yeah but wouldnt when i simplifed using e^()=e^() +e^K wouldnt this k=0 add a 1 and mess up the solution......
     
  18. May 13, 2015 #17
    e^K = Cte an other constant
    After you write that, ignore K, it's all about Cte
     
  19. May 13, 2015 #18
    So i just resolved the problem a different way and at the end I got y=(x^2+y^2)c. These are more circles that are orthogonal
     
  20. May 13, 2015 #19

    HallsofIvy

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    Science Advisor

    "More circles that are orthogonal" to what?
     
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