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Intro. to Differential Equations

  1. Jan 10, 2006 #76


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    another topic that is often omitted, but that arnol'd clarifies, is that of the domain of the solution. i.e. if the manifold where then vector field is defiend is compact, then the domain of the solution is all of R, so one gets a 1 parameter flow on the whole manifold, i.e. any point can be flowed along a solution curve for time t, whatever t is. this adds much to the geometry of the subject.
  2. Jan 27, 2006 #77
    guys,was just reading diff.eqns by ross.
    uniqueness of an 1-d d.eqn of the form
    y'=f(x,y) is said to be guaranteed when
    i>partial der.of f(x,y) w.r.t y is a continuous of x & y over the same domain D over which f(x,y) is defined & is continuous.
    can somebody explain how this condition guarantees uniqueness of the soln??
    Last edited: Jan 27, 2006
  3. Jan 27, 2006 #78


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    In order that the differential equation, dy/dx= f(x,y) have a unique solution (in some region around x0) satisfying y(x0)= y0 it is sufficient but not necessary that f be differentiable with respect to y in some region around (x0,y0. It is sufficient that f(x,y) be "Lipschitz" in y: If (x, y1) and (x,y2) are two points in that region, then |f(x,y1)- f(x,y2)|< C|y1- y2|. If a function is differentiable in a region, then you can use the mean value theorem to show that it is Lipschitz in that region but there exist functions that are Lipschitz in a region but not differentiable there. Most elementary texts use the simpler "differentiable" as a necessary condition and may or may not point out that it is not sufficient.

    Picard's proof of the existence and uniqueness of such a solution is long and deep. I won't post it here, but this is a link to a detailed explanation:

    http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/$file/Picard.PDF [Broken]

    Essentially, you replace the differential equation by the corresponding integral equation (dy/dx= f(x,y) if and only if [itex]y= \int f(x, y(x)) dx[/itex]) then show that the integral operator defined in that equation satisfies the hypotheses of Banach's fixed point theorem.
    Last edited by a moderator: May 2, 2017
  4. Feb 20, 2006 #79
    sorry for responding after such along time. 'was sick & hence away from my institute.
    thanx halllsofivy, but i did not ask how to prove that continuity of p.der of f wrt y was a sufficient condition, rather i wanted to know if it is possible to EXPLAIN this requirement in simpler terms !!
    let me explain,existence of a soln of the problem of the type given by me above is guaranteed if there f(x,y) is a cont fn of x& y in some domain D.this can be explained thus:: if f(x,y) was not cont, it wud not have been possible to integrate f(x,y) over the domain D (which is after all the method by which we find out the soln).
    what i'm seeking is some explanation of this nature!!
    thanx in advance!!
  5. Feb 20, 2006 #80


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    ??Why not? The fact that a function is not continuous does not mean it is not integrable. Even if that were true that would be showing that continuity is a necessary condition, not a sufficient condition, which is what by "solution is guaranteed". If that's the kind of "explanation" you are looking for, I surely can't help you!
  6. Aug 29, 2006 #81
    Just Wondering How....

    Hey Guys,

    I have learned that Mars and Earth closest encounter so far happened August 27... and it is estimated to commence again by year 2287 if im not mistaken.... I am just wondering how the calculations were made employing the principles of Differential Equations? Im really curious 'bout the accuracy or exactness of the date or just the exact year only...

    Hoping for your prompt response.

  7. Dec 4, 2006 #82
    hi to everyone
    I am a new user and happy to find this forum.so i'm looking for a way to solve these equations by using delphi programming language.i am a physics student and have "use computer in physics" in this course.
    thank you
  8. Feb 21, 2007 #83
    i just found this useful topic to use in my DE class. does anyone know where i could find solutions (the ones located on the first page of this topic)? they are all too old or not available.
  9. Apr 5, 2007 #84
    I tink what hawaii meant was to find the solution to the diff. eqn [tex]3x^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 [/tex] at the point xo = 0. After looking through the "Series solution of 2nd Order Linear Equations" in thread #47 written by ExtravagantDreams, i realize that the above diff. eqn can be solved easily. The solution will be of the form y = [tex]\sum_{n=0}^\infty a_n(x - x_0)^n [/tex].
    Since xo = 0, y = [tex]\sum_{n=0}^\infty {a_n x^n} [/tex].
    y' = [tex]\sum_{n=0}^\infty {a_n nx^{n-1}} [/tex].
    y'' = [tex]\sum_{n=0}^\infty {a_n n(n-1)x^{n-2}} [/tex].
    Now, substitute these expressions into the original diff. eqn:
    [tex]3x^2 \sum_{n=0}^\infty {a_n n(n-1)x^{n-2}} - x \sum_{n=0}^\infty {a_n nx^{n-1}} + \sum_{n=0}^\infty {a_n x^n} = 0 [/tex]
    Factor in the external x terms:
    [tex]\sum_{n=0}^\infty {3a_n n(n-1)x^n} - \sum_{n=0}^\infty {a_n nx^n} + \sum_{n=0}^\infty {a_n x^n} = 0 [/tex]
    Combining all the terms since they already have the same degree and same starting pt:
    [tex]\sum_{n=0}^\infty {[3n(n-1) - n + 1]a_n x^n} = 0 [/tex]
    So finally we arrive at 3n² - 4n + 1 = 0 and an = 0. We get 2 values of n, n=1 and 1/3. How do i proceed from here then??

    Any good help would be appreciated :)
    Last edited: Apr 5, 2007
  10. Apr 5, 2007 #85
    This diff. eqn is of the form [tex] a \frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + d = f(t) [/tex] where a,b,c and d are numeric constants. The solution to this is the same as that for 2nd order linear diff. eqns with constant coefficients, provided that f(t) = 0.
    So we have [tex]\frac{d^3x}{dt^3} - 2\frac{d^2x}{dt^2} + \frac{dx}{dt} = 0 [/tex].
    The 'kernel' or characteristic equation is in fact:
    r³ - 2r² + r = 0
    r(r²-2r+1) = 0
    r(r-1)² = 0
    r= 0, 1 (repeated)
    The general solution will be [tex]x = (c_1t + c_2)e^t + c_3 [/tex] where c1, c2 and c3 are constants of integration.

    Correct me if i'm wrong as i'm still new to differential equations :)
  11. Aug 27, 2008 #86
    Last edited by a moderator: Apr 23, 2017
  12. Sep 25, 2008 #87
    Hello Sir,
    I m a student of high energy particle physics. Sir i need the solution manual of Differential equations by S.Balachandra Rao. S.B.Rao is the professor in a college of Banglore, India. Sir please if u can do me a favor, plz give me the solution manual of this book. I shall be very very thankful to u. U can email me on this id lost_somewhere@live.com.

    Thank U.....
  13. Oct 7, 2008 #88
    What does

    [tex] \frac {d} {dt}[\mu (t)y] = \mu (t) g(t)[/tex]

    mean ? You see I don't have a copy of B&D. )
    Last edited: Oct 7, 2008
  14. Nov 3, 2008 #89
    hello, I'm also taking a class on ODE but i have a problem -i use An Intro course in Diff. eq.'s by Zill - that i get a nonsense result here is the eq:

    sin3x + 2y(cos3x)^3 = 0 (here ^ is to raise a power.how are u raising powers?)

    the last result i get which is nonsense ofcourse is: y^2 = -1/6(cos3x)^2. another result includes tan3x but is still negative.

    so y^2 is negative which is impossible. is the result right? I think there's a problem with the D.E. given.

    hope u can help. thx
  15. Dec 24, 2009 #90
    I had to solve a first-order nonlinear ODE which led me to a this equation.how can I find the solution for y?
  16. Dec 24, 2009 #91


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    Where are the derivatives, e.g., y', or differentials?
  17. Dec 25, 2009 #92


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    "Lambert's W function", W(x), is defined as the inverse function to f(x)= xex. Taking the W function of both sides gives y= W(f(x)).
  18. Dec 25, 2009 #93
    So the question should be solved numerically using the Lambert's W function? I mean that can't we then have a function in the form: y=f(x)? or we can no more go further than the Lambert's W function?
  19. Jan 17, 2011 #94


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    i need an insight and understanding of asymptotic behaviour as applied to singular cauchy problem....anyone can comment.........
    Last edited by a moderator: May 5, 2017
  20. May 7, 2012 #95
    good job
  21. Jul 3, 2012 #96
    Good night,

    Last week I begun to study differential equations by my own and first saw ODE's of separable variables. I've learned very well what they are and how to find constant and non-constant solutions. But something extremely trivial is boring me: I can't figure out why some ODE is or is not of separable variable. For example, I know that an ODE of s.v. is an ODE of the type

    [; \frac{dx}{dt} = g(t)h(x) ;]​

    but I simply cannot say why

    [; \frac{dy}{dx}=\frac{y}{x} ;]​

    is and ODE of s.v. and why

    [; \frac{dy}{dx}=\frac{x+y}{x^2 +1} ;]​

    is not.

    I know this is very trivial and I am missing something, but I don't know what. Can you help me, please? :-)


    Ps.: sorry for my lousy English.
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