MHB Existence of solution of a linear differential equation

[mathmari]

Hey!

I want to check if we can always find a solution of a linear differential equation of first order in the polynomial ring $F[z]$.

I have done the following:

The general linear differential equation of first order is $$ax'(z)+bx(z)=y(z)$$ where $x,y \in F[z]$.

Or is it possible that $y \notin F[z]$ ?

We multiply the equation by the integrating factor $\mu (z)$ and we get $$a\mu (z) x'(z)+b\mu (z) x(z)=\mu (z) y(z) \Rightarrow \mu (z) x'(z)+\frac{b}{a} \mu (z) x(z)=\frac{1}{a} \mu (z) y(z) \tag 1$$

We assume that $\frac{b}{a} \mu (z)=\mu'(z)$, so we have that $$\mu (z) x'(z)+\mu'(z) x(z)=(\mu (z) x(z))'$$

Replacing this at $(1)$ we get $$(\mu (z) x(z))'=\frac{1}{a} \mu (z) y(z)$$

Integrating the last relation we get $$\int (\mu (z) x(z))' dz=\int \frac{1}{a} \mu (z) y(z)dz \Rightarrow \mu (z) x(z)=\int \frac{1}{a} \mu (z) y(z)dz+c$$

We have that $\mu (z), y(z) \in F[z]$ so $\mu (z) y(z) \in F[z]$ and also $\int \mu (z) y(z)dz \in F[z]$.

But since not every fraction of polynomials is a polynomial we cannot divide by $\mu (z) $ at the equation $\mu (z) x(z)=\int \frac{1}{a} \mu (z) y(z)dz+c$ over $F[z]$.

So, there is not always a solution in the polynomial ring. Is everything correct? (Wondering)

 

[mathmari]

A correction: I have to find out if there is a solution of a linear differential equation of first order in the polynomial ring $\mathbb{C}[z]$. I tried it again and did the following:

• $\textbf{Case 1.}$
  $a=0, b \neq 0$
  
  Then we have $bx(z)=y(z)$.
  So, the solution is $$x(z)=\frac{1}{b}y(z) \in \mathbb{C}[z]$$
  
• $\textbf{Case 2.}$
  $a\neq 0, b=0$
  
  Then we have $ax'(z)=y(z)$.
  
  $$x'(z)=\frac{1}{a}y(z) \Rightarrow x'(z)=\frac{1}{a}\sum_{k=0}^nc_nz^k$$
  
  Then the solution is $$x(z)=\frac{1}{a}\sum_{k=0}^n c_k \int z^k dz+c$$
  So, $$x(z)=\frac{1}{a}\sum_{k=0}^n \frac{c_k}{k+1}z^{k+1}+c$$
  
• $\textbf{Case 3.}$
  $a=0, b=0$
  
  Then we have $0=y(z)$.
  
  If $y(z)=0$ then we have infinitely many solutions.
  If $y(z) \neq 0$ then we have no solution.
  
• $\textbf{Case 4.}$
  $a \neq 0, b \neq 0$
  
  then we have $ax'(z)+bx(z)=y(z) \Rightarrow x'(z)+\frac{b}{a}x(z)=\frac{1}{a}y(z)$.
  
  Let $y=\sum_{k=0}^n c_kz^k$ and $x(z)=\sum_{k=0}^n d_kz^k$, so $x'(z)=\sum_{k=1}^n kd_kz^{k-1}$.
  
  Then we have $$\sum_{k=1}^nkd_kz^{k-1}+\frac{b}{a}\sum_{k=0}^nd_kz^k=\frac{1}{a}\sum_{k=0}^nc_kz^k$$
  
  Comparing the coefficients we have $$\left\{\begin{matrix}
  \frac{b}{a}d_n=\frac{1}{a}c_n \Rightarrow d_n=\frac{1}{b}c_n \\
  kd_k+\frac{b}{a}d_{k-1}=\frac{1}{a}c_{k-1}, \ \ k=1, \dots , n-1
  \end{matrix}\right.$$
  
  Solving this recursive relation we get the solution $x(z)$.

Is everything correct? (Wondering)