Differential equations - Relation between the number of solutions and the order The case $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$: 
I want to show that in the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$ each differential equation has a solution. 
I have done the following: 
Each element of the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$ is of the form $\displaystyle{\sum_{k=1}^n α_kz^{d_k}e^{β_kz}}$. 
A differential equation in this ring is of the form $$Ly = \sum_{k=0}^m \alpha_k y^{(k)}(z)=\sum_{l=1}^n C_lz^{d_l} e^{\beta_l z} , \ \ \alpha_k , \beta_l \in \mathbb{C} \ \ \ \ (*)$$  
At the linear differential equations we can apply the superposition principle. That means that we can split the problem $(*)$ into the subproblems $$Ly(z)=0, \ \ Ly(z)=C_lz^{d_l} e^{\beta_l z}, \ \ l=1, 2, \dots , n$$ so into an homogeneous and $n$ non-homogeneous equations. We solve these equations and then we add the solution of the homogeneous $y_{H}(z)$ and the solutions $y_{p_i}(z)$of the $n$ non-homogeneous equations. 
So we get the general solution of the equation $(*)$, which is $$y(z)=y_{H}(z)+\sum_{l=1}^n y_{p_i}(z)$$ 
To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=0$$ we find the characteristic equation and its eigenvalues $\lambda_1, \dots , \lambda_m$. 


*

*If $\lambda_1, \dots , \lambda_m$ are eigenvalues of multiplicity $1$, then the solution of $Ly(z)=0$ is $$y_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$ 

*If $\lambda_i$ is an eigenvalues of multiplicity $M>1$, then the $$e^{\lambda_i z}, ze^{\lambda_i z}, z^2e^{\lambda_i z}, \dots , z^{M-1}e^{\lambda_i z}$$ are $M$ linear independent solutions of $Ly(z)=0$. 


To solve the equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=C_lz^{d_l} e^{\beta_l z} \ \ \ \ \ (**)$$ we do the following:  


*

*If $\beta_l$ is not one of the eigenvalues: 
We suppose that the solution is in the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$, so it is of the form $$y(z)=e^{\beta_l z} x(z)$$ 
So $$y^{(k)}(z)=\sum_{j=0}^k \binom{k}{j}\beta_l^j e^{\beta_l z}x^{(k-j)}(z)$$ 
Substituting in $(**)$ we get the following equation, the order of which is the same as the order of $(**)$, $$\sum_{k=0}^m \beta_k x^{(k)}(z)=C_lz^{d_l}$$  
The solution of the above equation will be polynomials. 
The first non-zero term $\beta_k$ determines the degree of $x$. 
For example if $\beta_o \neq 0$ then the solution will have degree at most $l$, and will be of the form $$x(z)=\gamma_l z^l +\dots +\gamma_0$$ 
Then $$x'(z)=l\gamma_l z^{l-1}+\dots +\gamma_1 \\ \dots \\ x^{(l)}(z)=l!\gamma_l$$ 
Then $$\sum_{k=0}^m \beta_k x^{(k)}(z)=C_lz^{d_l}  \Rightarrow \beta_o\gamma_kz^{d_l}+(\beta_0\gamma_{l-1}+l\gamma_la_1)z^{d_l-1}+ \dots =C_lz^{d_l} $$
So it must stand $$\beta_0\gamma_l=C_l \Rightarrow \gamma_l=\frac{C_l}{\beta_0} \\ \beta_0\gamma_{l-1}+l\gamma_la_1=0 \\ \dots $$ 
So to solve the differential equation we have to solve the above system. 

We write this in the form of matrix. Can we be sure that we will find an unique solution?
In this ring the number of solutions is equal to the number of the order, right? How can we justify that?



*

*If $\beta_l$ is one of the eigenvalues $\lambda_i$, and the multiplicity of $\lambda_i$ is $L$: 
We suppose that the solution is in the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$, so it is of the form $$y(z)=z^L e^{\beta_l z} x(z)=e^{\beta_l z}\tilde{x}(z)$$ and we continue as in the previous case.  

Is everything correct? 
Could I improve something at the formulation? 


The case $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$: 
I want to check when in the ring $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$ a differential equation has a solution. 
I have done the following: 
Each element of the ring $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$ is of the form $\displaystyle{\sum_{k=1}^n α_ke^{β_kz}}$. 
A differential equation in this ring is of the form $$Ly = \sum_{k=0}^m \alpha_k y^{(k)}(z)=\sum_{l=1}^n C_l e^{\beta_l z} , \ \ \alpha_k , \beta_l \in \mathbb{C} \ \ \ \ (*)$$  
At the linear differential equations we can apply the superposition principle. That means that we can split the problem $(*)$ into the subproblems $$Ly(z)=0, \ \ Ly(z)=C_l e^{\beta_l z}, \ \ l=1, 2, \dots , n$$ so into an homogeneous and $n$ non-homogeneous equations. We solve these equations and then we add the solution of the homogeneous $y_{H}(z)$ and the solutions $y_{p_i}(z)$of the $n$ non-homogeneous equations. 
So we get the general solution of the equation $(*)$, which is $$y(z)=y_{H}(z)+\sum_{l=1}^n y_{p_i}(z)$$ 
To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=0$$ we find the characteristic equation and its eigenvalues $\lambda_1, \dots , \lambda_m$. 


*

*If $\lambda_1, \dots , \lambda_m$ are eigenvalues of multiplicity $1$, then the solution of $Ly(z)=0$ is $$y_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$ 

*If $\lambda_i$ is an eigenvalues of multiplicity $M>1$, then the $$e^{\lambda_i z}, ze^{\lambda_i z}, z^2e^{\lambda_i z}, \dots , z^{M-1}e^{\lambda_i z}$$ are $M$ linear independent solutions of $Ly(z)=0$. But in this case the solutions are not in $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$. So we reject this case. 


To solve the equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=C_l e^{\beta_l z} \ \ \ \ \ (**)$$ we do the following:  


*

*If $\beta_l$ is not one of the eigenvalues: 
We suppose that the solution is in the ring $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$, so it is of the form $$y(z)=A_l e^{\beta_l z}$$ 
So $$y^{(k)}(z)=\beta_l^k A_l e^{\beta_l z}$$ 
Substituting in $(**)$ we get the following equation, the order of which is the same as the order of $(**)$, $$\sum_{k=0}^m \alpha_k \beta_l^k A_l=C_l \Rightarrow A_l=\frac{C_l}{\sum_{k=0}^m \alpha_k \beta_l^k}$$  
The solution of the above equation will be exponentials. 

*If $\beta_l$ is one of the eigenvalues $\lambda_i$, and the multiplicity of $\lambda_i$ is $L$: 
Then the solution is of the form $$y(z)=B_lz^Le^{\beta_l z}$$ but this is not an element of the ring. So we reject this case.  

The case $\mathbb{C}[z]$: 
I want to check when in the ring $\mathbb{C}[z]$ a differential equation has a solution. 
I have done the following: 
Each element of the ring $\mathbb{C}[z]$ is of the form $\displaystyle{\sum_{k=1}^n α_kz^{k}}$. 
A differential equation in this ring is of the form $$Ly = \sum_{k=0}^m \alpha_k y^{(k)}(z)=\sum_{l=1}^n C_lz^{l}  , \ \ \alpha_k  \in \mathbb{C} \ \ \ \ (*)$$  
At the linear differential equations we can apply the superposition principle. That means that we can split the problem $(*)$ into the subproblems $$Ly(z)=0, \ \ Ly(z)=C_lz^{l} , \ \ l=1, 2, \dots , n$$ so into an homogeneous and $n$ non-homogeneous equations. We solve these equations and then we add the solution of the homogeneous $y_{H}(z)$ and the solutions $y_{p_i}(z)$of the $n$ non-homogeneous equations. 
So we get the general solution of the equation $(*)$, which is $$y(z)=y_{H}(z)+\sum_{l=1}^n y_{p_i}(z)$$ 
To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=0$$ we find the characteristic equation and its eigenvalues $\lambda_1, \dots , \lambda_m$. 


*

*If $\lambda_1, \dots , \lambda_m$ are eigenvalues of multiplicity $1$, then the solution of $Ly(z)=0$ is $$y_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$ 
But then $y_{H} (z) \notin \mathbb{C}[z]$. So we reject this case. 

*If $\lambda_i$ is an eigenvalues of multiplicity $M>1$, then the $$e^{\lambda_i z}, ze^{\lambda_i z}, z^2e^{\lambda_i z}, \dots , z^{M-1}e^{\lambda_i z}$$ are $M$ linear independent solutions of $Ly(z)=0$. 
The solutons are in $\mathbb{C}[z]$ only if $\lambda_i=0$. 


To solve the equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=C_lz^{l}  \ \ \ \ \ (**)$$ we do the following:  


*

*The multiplicity of $\lambda_i=0$ is $L$: 
We suppose that the solution is in the ring $\mathbb{C}[z]$, so it is of the form $$y(z)=z^L x(z)=\tilde{x}(z).$$
So $$y^{(k)}(z)=\tilde{x}^{(k)}(z)$$ 
Substituting in $(**)$ we get the following equation, the order of which is the same as the order of $(**)$, $$\sum_{k=0}^m \alpha_k  \tilde{x}^{(k)}(z)=C_lz^{l}$$  
The solution of the above equation will be polynomials. 
The first non-zero term $\alpha_k$ determines the degree of $\tilde{x}$. 
For example if $\alpha_0 \neq 0$ then the solution will have degree at most $l$, so $x$ will be at most $l-L$, and will be of the form $$x(z)=\gamma_l z^{l-L} +\dots +\gamma_0\Rightarrow \tilde{x}(z)=z^Lx(z)=\gamma_l z^{l} +\dots +\gamma_0 z^L$$ 
Then $$\tilde{x}'(z)=l\gamma_l z^{l-1}+\dots +L\gamma_0z^{L-1} \\ \dots \\ \tilde{x}^{(l)}(z)=l!\gamma_l$$ 
Then $$\sum_{k=0}^m \alpha_k \tilde{x}^{(k)}(z)=C_lz^{d_l}  \Rightarrow \alpha_0\gamma_kz^{d_l}+ \dots =C_lz^{d_l} $$
So it must stand $$\alpha_0\gamma_l=C_l \Rightarrow \gamma_l=\frac{C_l}{\alpha_0} \\ \dots $$ 
So to solve the differential equation we have to solve the above system. 
So, a differential equation has a solution when the eigenvalues are $0$ and of multiplicity $>1$. 
 A: For the case $\mathbb{C}[z,e^{\lambda z} | \lambda \in \mathbb{C}]$ and when $\beta_l$ is not an eigenvalue, note that $\beta_0 = p(\beta_l) := \sum_{k=0}^m \alpha_k \beta_l^k$ ($\beta_0$ has a different meaning than $\beta_l$, which needs to be renamed IMO), where $p(\lambda)$ is the characteristic polynomial. Since $\beta_l$ is not an eigenvalue, $\beta_0 \neq 0$ and $\deg x(z) = d_l$. So the coefficient matrix for the unknowns $\gamma_i$, is a $(d_l+1)\times(d_l+1)$ lower triangular matrix that has $\beta_0$ on the diagonals (the matrix came from the $d_l+1$ linear equations as a result of polynomial equality). Since the coefficient matrix is nonsingular, the values $\gamma_i$ can be determined uniquely.
In the case $\beta_l = \lambda_i$, degree of $x(z)$ is still $d_l$. Therefore, we have $L + d_l + 1$ equations, where first $L$ equations are already zero, because the multiplicity of $\lambda_i$ is $L$. So the first nonzero coefficient is $\beta_L = c p^{(L)}(\beta_l)$ for some nonzero constant $c$. (Note $p^{(k)}(\beta_l) = 0$ for $k=0,1,\dots,L-1$). The remaining steps are the same as above.
(By the way, you need to select $y(z) = z^{L-1} e^{\beta_l z} x(z)$)
For the case $\mathbb{C}[e^{\lambda z} | \lambda \in \mathbb{C}]$ I think you need to explain why $\sum_{k=0}^m \alpha_k \beta^k \neq 0$ before dividing.
For the case $\mathbb{C}[z]$, just select $\beta_l = 0$ in the first case. First consider the case when the characteristic polynomial has no $0$ roots. There are no homogenous solutions in the ring in this case. For nonhomogenous case $L y = z^{d_l}$, take the solution $y=\sum_{i=0}^{d_l} \gamma_i z^i$ and solve for $\gamma_i$ as in the previous case. If the characteristic polynomial has a $0$ root with multiplicity $M$, there are $M$ polynomial solutions for the homogenous equation, $1, z, \dots, z^{M-1}$. Now select $y=z^M \sum_{i=0}^{d_l} \gamma_i z^i$ and solve $M+d_l+1$ equations, where the first $M$ are already zero.
For all cases, the solution is unique for given initial conditions.
