0
$\begingroup$

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?

Thanks in advance!!

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Yess it follows ...but what can we say about maximum number of linearly independent solutions in the solution set?are they only n+1 in number as listed by me in question or it can even reach this figure? $\endgroup$ Dec 26, 2022 at 18:11
  • $\begingroup$ I will edit the answer. $\endgroup$
    – Gerd
    Dec 26, 2022 at 21:15
  • $\begingroup$ Thank you so much for help $\endgroup$ Dec 28, 2022 at 1:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .