Every $n^{th}$ order non-homogenous linear differential equation has exactly $'n+1'$ linearly independent solutions

Prove or disprove:

Every $$n^{th}$$ order non-homogenous linear differential equation has exactly $$'n+1'$$ linearly independent solutions.

I know that if $$y_1,y_2,...y_n$$ are linearly independent complimentary functions(i.e solutions of corresponding homogenous equation) and $$y_p$$ is a particular integral then $$y_p,y_1+y_p,y_2+y_p,...y_n+y_p$$ are $$n+1$$ solutions of non-homogenous linear differential equation.Are these soltions linearly independent?If it is so then we are done.I literally needs your help as i cannot proceed from here.Further i want to know if this statement is not true then how many solutions of a non-homogenous linear differential equation are linearly independent?

I think these solutions are linearly independent: Let $$0=\alpha_0y_p + \alpha_1 (y_1+y_p)+ \dots +\alpha_n (y_n+y_p) = \left(\sum_{k=0}^n \alpha_k\right) y_p + \sum_{k=1}^n \alpha_ky_k.$$ The solution set of the homogenous equation is exactly the span of $$\{y_1,\dots,y_n\}$$. Since $$y_p$$ is not a solution of the homogenous equation we can conclude $$\sum_{k=0}^n \alpha_k =0$$. Then $$\alpha_1=\dots=\alpha_n=0$$ follows (since $$y_1,\dots y_n$$ are linearly independent), hence also $$\alpha_0=\sum_{k=0}^n \alpha_k =0$$.
Edit: Let $$z_1,\dots, z_{n+2}$$ be any solutions of the inhomogenous equation. Set $$w_1:=z_2-z_1, ~ w_2:=z_3-z_1, ~ \dots, ~ w_{n+1}:=z_{n+2}-z_1.$$ Then $$w_1,\dots, w_{n+1}$$ are $$n+1$$ solutions of the homogenous equation, hence linearly dependent. Thus there is a nontrivial linear combination $$0=\sum_{k=1}^{n+1} \beta_k (z_{k+1}-z_1) = \sum_{k=1}^{n+1} \beta_k z_{k+1} - \left(\sum_{k=1}^{n+1} \beta_k \right)z_1.$$ Thus $$z_1,\dots, z_{n+2}$$ are linearly dependent.