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Working on Constructing Equations

  1. Aug 14, 2004 #1
    I come across some problems,

    I'm working on finding Differential Equations for n-parameter Family of solutions. However the problem is constructing the n-parameter family equation to work with to find a Differential Equation for the particular n-parameter family equation.

    Here's the problem,

    Find a differential equation whose solution is

    18. A family of circles of fixed radii and centers on the x axis.

    Now i know that a family of circles with center at origin is

    [tex] x^2 + y^2 = r^2, r > 0 [/tex]

    However I do not know how to form an equation such that a family of circles of fixed radii, centers on the x axis. I find that the fixed radii can be represented by a constant?

    The answer for 18 is

    [tex] (yy')^2 + y^2 = a^2 [/tex]

    That is the differential equation, however I do not know how to get the differential equation if I can't construct a family of circles of fixed radii that centers on the x axis.

    Here is another problem

    19. A family of circles of variable radii, centers on the x axis and passing through the origin.

    There is only one problem such as problem 21 that gives the equation to you, so you don't have to construct it.

    21. A family of circles with centers in the xy-plane and of variable radii. Then it gives a Hint: Write the equation of the family as

    [tex] x^2 + y^2 - 2c_1x - 2c_2y + 2c_3 = 0 [/tex]

    I read the problems, and I get an idea of what it wants but, the construction of an equation is difficult.
  2. jcsd
  3. Aug 16, 2004 #2


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    The general equation for a circle of radius r with center at (a,b) is
    [tex](x-a)^2+ (y-b)^2= r^2 [/tex]

    If the center is on the x axis, then b= 0 so the equation for "a family of circles, of fixed radii, centers on the x axis" is
    [tex](x-a)^2+ y^2= r^2[/tex].
    You want a to take on all possible values but r is a "given" constant.

    Since you want a differential equation, differentiate!
    2(x-a)+ 2y dy/dx= 0 or x-a = yy'. IF the problem had been "a family of circles with center at (a,0), differing radii, you would be done now but since it is a that is differing for different circles in the family, you need to eliminate a.
    From the original equation, (x-a)2= r2- y2 so the differential equation can be written r2- y2= (yy')2 which is, of course, the same as (yy')2+ y2= r2. (The answer in the book is using "a" as the radius rather than "r".)

    If the center is on the x- axis at, say, (a,0), and it passes through the origin then r= a. It's equation is (x- a)2+ y2= a2.
    Differentiating, 2(x-a)+ 2yy'= 0 or x-a= yy' so (x-a)2= (yy')2. Using (x-a)2= a2- y2 to get
    (yy')= a2- y2 isn't quite enough since you still have "a" in the equation. (I'll leave the rest of it to you.)

    There are 3 constants (this is a family with "3 degrees of freedom") so you will need a third order differential equation.
    differentiating, 2x+ 2yy'- 2c1- 2c2y'= 0.
    Differentiating again, 2+ 2yy"+ 2y'2- 2c2y"= 0.
    Once again: 2yy'''+ 4y'y"- 2c2y'''= 0.

    You should be able to use those four equations (including the original equation of the family) to eliminate the three constants.
    Last edited by a moderator: Aug 16, 2004
  4. Aug 17, 2004 #3
    Thank you so much! I think I can get the rest of problems now.

    There are many more problems that deal with Family of parabolas and such. Which should be alot easier, since the equation of the parabola is alot more simple. I should of known how to construct differentiable equations of circles being that I had the equation

    [tex] (x-h)^2 + (y-k)^2 = r^2 [/tex] with center at (h,k)

    in front of my eyes the whole time.
  5. Feb 16, 2006 #4
    Plzz help!!

    I was reading this and wat abt the case where we have to find differential eq. of circles passing through origin?
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