Solutions to Linear Diff. Eq. of 1st Order in a Ring?

[mathmari]

Hey!

When we want to have solutions of a linear differential equation of first order $$ax'(z)+bx(z)=y(z)$$ in a ring $R$, does $y$ have to be an element of the ring?

Or is it possible that $y$ is a function that does not belong to $R$ but the solution of the differential equation is in $R$ ? (Wondering)

 

[I like Serena]

Hey!

When we want to have solutions of a linear differential equation of first order $$ax'(z)+bx(z)=y(z)$$ in a ring $R$, does $y$ have to be an element of the ring?

Or is it possible that $y$ is a function that does not belong to $R$ but the solution of the differential equation is in $R$ ? (Wondering)

Hey mathmari! (Wave)

Wouldn't $a,b,x(z),x'(z)$ all be elements of $R$? (Wondering)
Because that would imply that $y(z)$ is also an element of $R$.

 

[mathmari]

Wouldn't $a,b,x(z),x'(z)$ all be elements of $R$? (Wondering)

Yes.

Because that would imply that $y(z)$ is also an element of $R$.

I see... Thanks a lot! (Mmm)

 

[mathmari]

So, we have the following:

A differential equation $$\alpha_ny^{(n)}+\dots +\alpha_0 y=F$$ with constants coefficients $\alpha_i \in R$, $i=0, \dots , n$ and $\alpha_n \neq 0$ has solutions in the ring $R$ if $F\in R$. right?

Is it a necessary but not sufficient condition ? (Wondering) Is the proof of the above statement the following? (Wondering)

We suppose that the solution of the differential equation belongs to the ring, i.e., $y \in R$. Then also each of its derivatives belongs to the ring, i.e., $y^{(i)} \in R$, $i=1, \dots , n$.
We also have that $\alpha_i \in R$, $i=0, \dots , n$.
Then $\alpha_ny^{(n)}+\dots +\alpha_0 y \in R \Rightarrow F\in R$.