# Find simplest differential equation

1. Nov 24, 2009

### 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.

2. Nov 25, 2009

### 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.

3. Nov 27, 2009

### matematikawan

ex 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

4. Nov 30, 2009

### 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.

5. Dec 1, 2009

### 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.

6. Dec 1, 2009

### 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 .

7. Dec 1, 2009

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

Last edited: Dec 1, 2009
8. Dec 1, 2009

### Mute

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.

9. Dec 2, 2009

### matematikawan

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