Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

ODEs with variable coefficients

  1. May 14, 2010 #1
    Can anyone help me to get the general solution of the linear partial differential equations with variable coefficients of any order?
  2. jcsd
  3. May 14, 2010 #2


    User Avatar
    Science Advisor

    No one knows how to do that. PDEs are quite complicated.
  4. May 14, 2010 #3
    can anyone help me to solve this ODE:
    where: A(x) and M(x) are variable coefficients
    Y''''(x,t) 4th derivative with respect to x.
    w''(x,t) 2nd derivative with respect to t.
  5. May 14, 2010 #4


    User Avatar
    Science Advisor
    Gold Member

    That is a PDE, not an ODE. What is A, what is M, what is Y, what is W? What are boundary/initial values? Unless you get specific, espcially regarding A and M, no one can help you.

  6. May 15, 2010 #5
    this PDE is correct. but when i use the seperation of variables principle to solve this equation, the resulting two equations are ODEs, which may as follows:
    A(x)Y''''(x,t)+M(x)Y''(x,t)=0 ........(1) (main equation)
    W''(t)-aW(t)=0 ..........(2)
    where: a is arbitrary constant. this may be solved easily.and the second ODE is:
    {A(x)/M(x)}F''''(x)-aF(x)=0 .........(3)
    this equation is linear ODE with variable coefficient.
    in which: A(x) and M(x) are variable coefficients along x-axis.
    a is arbitrary constant, from seperation of variable process.
    by solving equation(3) and compiling with the solution of equation(2), we will get the solution of equation(1). so, the problem is how we can solve the equation(3). boundary or initial conditions, i don't we need now to solve equation(3).
  7. May 15, 2010 #6


    User Avatar
    Science Advisor

    [tex]A(x)\frac{\partial^4 Y}{\partial x^4}+ M(x)\frac{\partial^2M}{\partial t^2}= 0[/tex]
    Not only do you have two independent variables, x and t, you have two dependent variables, Y and M. One equation is not sufficient to solve for two unknown functions.
    Last edited by a moderator: Jun 2, 2010
  8. May 16, 2010 #7
    What you seemed to want was:

    [itex]A(x)\frac{\partial^4 Y}{\partial x^4}+M(x)\frac{\partial^2 Y}{\partial t^2}=0[/itex]

    and by letting [itex]Y(x,t)=F(x)W(t)[/itex] and separation constant equal to a, you get one ODE in t and another ODE in x. You then want to know how to solve the ODE in x:

    [itex]\frac{d^4 F}{dx^4}=b(x)F[/itex]

    where [itex]b(x)=\frac{aM(x)}{A(x)}[/itex]

    Here's what I'd do with this and every other problem like it: First just try it in Mathematica or Wolfram Alpha:

    DSolve[y''''[x] == b[x] y[x], y, x]

    who knows, might get an answer and then that answer could conceivably help you derive an analytical method to obtain it. However in this case, Mathematica can't come up with one. Doesn't mean some nice method can't find it but rather, just Mathematica. However if A(x) is a constant and M(x) is a polynomial, then looks like you could solve it via simple power series. Additionally if b(x) is analytic with power series [itex]b(x)=\sum c_n x^n[/itex] you could still conceivably solve it via power series but would need to form the Cauchy Product of the resulting series product. For example, the equation [itex]y''''-e^x y=0[/itex] is not too hard to solve this way for say x in (0,1). See "Intermediate Differential Equations" by Rainville for a (simple) example.
    Last edited: May 16, 2010
  9. May 21, 2010 #8
    i want to ask. what are the "differential forms"? and what are the distinct applications?
  10. Jun 2, 2010 #9
    Hi colleages. can you help me to solve the cubic equation below:
    2N(Ep-En)hp^3(x)-3[M(x,t)(Ep-En)-2NEnh]hp^2(x)-6Enh[M(x,t)+Nh]hp(x)+Enh^2[3M(x,t)+2Nh]=0 notice that all variables in the equation are dependent on x only, except M is dependent on x and t.
    En, Ep, N and h are constants.
    i know the solution when the equation dependent on one variable, but the problem is that M dependent on two variables x and t.
    please help me and i will be very grateful for you...
  11. Jun 12, 2010 #10

    U''''=4th partial derivative with respect to x only.
    V''=2nd partial derivative with respect to t only.
  12. Jun 18, 2010 #11
    please, my colleages, can tell me is the equation below is linear or nonlinear PDE?
    U''''=4th partial derivative with respect to x only.
    V''=2nd partial derivative with respect to t only.
  13. Jun 18, 2010 #12


    User Avatar
    Science Advisor

    Since you have no powers or more complicated functions of the dependent variables U and V, that is a linear equations. However, I will say once more- if you have two dependent variables to solve for, you will need two equations.
  14. Jun 18, 2010 #13
    So what about the equation I asked about in https://www.physicsforums.com/showthread.php?t=410984 [Broken], which is

    [tex]\Delta[/tex]u(x,y) = u(x,y)*f(x,y)

    Does this fall into the same category? It seems to me that there is only one unknown equation with two variables, however, does this count as two equations if f is separable?
    Last edited by a moderator: May 4, 2017
  15. Jun 19, 2010 #14
    Thank you very much. but is the equation ( EI(x,t)U''''(x,t)+M(x,t)V''(x,t)=0 )separable or not?How?if it isnot separable how can be solved?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook