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Find simplest differential equation

  1. Nov 24, 2009 #1
    How can I find the simplest differential equation given that I know the set of particular solutions? For example, take [tex]y_p=\{\ln(x), e^x\}[/tex], how can I find the simplest differential equation that has these solutions?

    Edit: It has to be homogeneous.
  2. jcsd
  3. Nov 25, 2009 #2
    If it has to be linear homogeneous, then for example

    (-x+x^2\ln(x))y''= (1+x^2\ln(x))y'-(x+1)y

    Formal way to find LODE is as follow.
    You can form the "general" solution as


    and then eliminate C_1,...C_n by successive differentiations.
  4. Nov 27, 2009 #3
    ex is a solution for y'=y.

    ln(x) is a solution for y'=x-1.

    Then [tex]y_p=\{\ln(x), e^x\}[/tex]

    satisfy the DE (y'-y)(y'-x-1)= 0 :biggrin:
  5. Nov 30, 2009 #4
    plug the solutions into a general form y''+p*y'+q*y=0 and you get two algebric equations which determines p and q.
  6. Dec 1, 2009 #5
    Actually how many different ways can we form DE that has solutions y=e^x and y= ln x ?

    Is my equation (y'-y)(y'-x-1)= 0 equivalent to (-x+x^2\ln(x))y''= (1+x^2\ln(x))y'-(x+1)y ?

    I know a first order nonlinear Riccati DE can be reduced to a linear 2nd order DE. But not particularly sure of this one. Is the degree of a DE has any role to play?
    Definition: The degree of a DE is the degree of the highest ordered derivative.
  7. Dec 1, 2009 #6
    There are uncountable different ways to form DE (it'll be different DEs) that has solutions y=e^x and y= ln x if we do not restrict ourselves by homogeneous linear DEs. For example,

    first order ODE
    x*(exp(x)-ln(x))*diff(y(x),x)+(-exp(x)*x+1)*y(x)+(-1+x*ln(x))*exp(x)=0 ,

    second order ODE
    -ln(x)*x^2*(ln(ln(x))*ln(x)-1)*diff(y(x),x)^2-y(x)*x*(ln(x)+1)*diff(y(x),x)+y(x)*ln(x)*x^2*(ln(ln(x))*ln(x)-1)*diff(y(x),`$`(x,2))+factor(y(x)^2*ln(y(x))*ln(x)+y(x)^2*ln(y(x)))=0 .
  8. Dec 1, 2009 #7
    Even just looking at the form (y' - y)(y' - x-1) = 0, there are any number of ways to form a differential equation with the given solutions, since Dn(ex) = ex, one can use (y(n) - y)(y' - x-1) = 0, for any positive integer n.
    Last edited: Dec 1, 2009
  9. Dec 1, 2009 #8


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    Homework Helper

    You have to be careful, though. You proposed DE has introduced new solutions: [itex]\exp[\omega_n x][/itex], where [itex]\omega_n[/itex] is an n-th root of 1. So while such a DE does have the desired solutions, it now has many more, which may not be desired.
  10. Dec 2, 2009 #9
    OK I will take note that 'simplest DE' means linear DE not the lowest order DE.
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