- #1
mathmari
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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)
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)