Find simplest differential equation

[epkid08]

How can I find the simplest differential equation given that I know the set of particular solutions? For example, take y_p=\{\ln(x), e^x\}, how can I find the simplest differential equation that has these solutions?

Edit: It has to be homogeneous.

 

[kosovtsov]

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

y=C_1y_1+...+C_ny_n

and then eliminate C_1,...C_n by successive differentiations.

 

[matematikawan]

e^x is a solution for y'=y.

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

Then y_p=\{\ln(x), e^x\}

satisfy the DE (y'-y)(y'-x^-1)= 0

 

[elibj123]

plug the solutions into a general form y''+p*y'+q*y=0 and you get two algebric equations which determines p and q.

 

[matematikawan]

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.

 

[kosovtsov]

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 .

 

[pbandjay]

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(e^x) = e^x, one can use (y⁽ⁿ⁾ - y)(y' - x^-1) = 0, for any positive integer n.

 

[Mute]

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(e^x) = e^x, one can use (y⁽ⁿ⁾ - y)(y' - x^-1) = 0, for any positive integer n.

You have to be careful, though. You proposed DE has introduced new solutions: \exp[\omega_n x], where \omega_n 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.

 

[matematikawan]

OK I will take note that 'simplest DE' means linear DE not the lowest order DE.