Check if there is a solution of the differential equation In my notes there is the following: 
$$y''+y'=1$$ 
The particular solution is a polynomial, say $f$. 
The solution of the homogeneous equation is $y_H=c_1+c_2e^{-x}$. 
Therefore, the solution of the above differential equation is $$y=c_1+c_2e^{-x}+f$$ 
We want to check if there is a solution in the ring $\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$. 
It is related to the Wroksian. 
$$\sum_{i,j}\alpha_{i,j}x^ie^{jx}=0 \Rightarrow \alpha_{i,j}=0, \forall i,j$$ 
Linear independence.  
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Could you explain to me why and how it is related to the Wronskian? Why do we need the linear independence to be able to check for solutions in the ring. Or have I understood it wrong?  
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EDIT: 
In this case we are working in the ring $\mathbb{C}[e^{\lambda x } \mid \lambda \in \mathbb{C}]$, and $f$ is not an element of the ring. So, in this ring the differential equation has no solutions, right? Can we generalize it? When we have any differential equation, can we answer if it has a solution in the ring or not? To do that, do we use the Wronskian, the linear independence? 
 A: Your question may be a bit unclear. Let $y_1=1$ and $y_2=e^{-x}$ be the solutions of the homogeneous equation $y''+y'=0$ that you found. Their Wronskian $\begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix}=\begin{vmatrix} 1 & e^{-x} \\ 0 & -e^{-x} \end{vmatrix}=-e^{-x}\not=0$, hence $y_1$ and $y_2$ form a fundamental set of solutions for the homogeneous equation. That is, every solution of the homogeneous equation will be of the form $c_1 y_1 + c_2 y_2=c_1  + c_2 e^{-x}$ for suitable $c_1$ and $c_2$ (which depend on the initial conditions). 
I do not follow what you are asking about $\sum_{i,j}\alpha_{i,j}x^ie^{jx}=0 \Rightarrow \gamma_{i,j}=0, \forall i,j$, this sentence seems incomplete and out of context. 
If you do find just one particular solution $f$ for the given equation 
 $y''+y'=1$, then all solutions of this equation will be $c_1 y_1 + c_2 y_2 +f$, i.e. $c_1  + c_2 e^{-x} +f$. 
You may find (or verify) that $f(x)=x$ is a solution (one particular solution) of $y''+y'=1$. You could find this solution regardless if you found $y_1$ and $y_2$ for the homogeneous equation, and unrelated to their Wronskian. But, if the Wronskian of $y_1$ and $y_2$ is non-zero (which is indeed the case for $y_1=1$ and $y_2=e^{-x}$) then you could express all solutions of the homogeneous equation as $c_1 y_1 + c_2 y_2=c_1  + c_2 e^{-x}$. This in turn implies that all solutions of the given non-homogeneous equation $y''+y'=1$ could be expressed as $c_1 y_1 + c_2 y_2 +f$, i.e. $c_1  + c_2 e^{-x} +x$ for this example. 
