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In recent ODE test I faced the following problem:

What is the minimal order of ODE $y^{(n)}+a_1y^{(n-1)}+a_2y^{(n-2)}+\dots + a_ny=0, a_i \in \mathbb{R}$, if his fundamental set of solutions contains functions $1, x^2, \sin x$ : 3, 4 or 5?

My answer was 3, because I thought that $1, x^2, \sin x$ are linear independent, linear homogenous ODE or order $n$ has $n$ linearly undefended solutions, and since therefore order of equation may not be lower than 3.

However, my answer was wrong. Can you please point out my mistake and provide correct answer and solution?

Thanks a lot for your help!

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3 Answers 3

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In order that $x^2$ be a solution, the characteristic polynomial must have $\lambda = 0$ as a zero with multiplicity (at least) $3$.

In order that $\sin x$ be a solution, you must have $\lambda =\pm i$ as complex conjugate zeros.

Hence, the minimal order is $5$: $$ x^{(5)} + x^{(3)} = 0. $$

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You are correct in your reasoning, that the order of the equation may not be lower than three. However not all of these solutions are first order solutions. For example, $\sin x$ is the solution to $\ddot y + y =0$ but not to any first order homogeneous autonomous equations with real coefficients.

The minimal order should be five, with a basis given by $\{1,x,x^2,\sin x,\cos x\}$. Let's see why:

First, examine the solution $y=x^2$. Plugging this in, we find that

$$2a_{n-2}+2a_{n-1}x+a_nx^2=0.$$

By the linear independence of $1,x,x^2$ it follows that $a_{n-2},a_{n-1},a_n=0$. Thus we already have the order as being at least third order. This follows from the more general result that if $p(x)$ is an $k$th order polynomial which is the solution to a homogeneous, autonomous linear differential equation, then the coefficients of the lowest $k+1$ derivatives (including the $0$th derivative) must be zero.

This actually takes care of our $y=1$ case, and suggests a solution of $y=x$. Thus we have at least four fundamental solutions, $1,x,x^2,\sin x$. There are two ways to proceed. One is to note that any equation for which $\sin x$ is a solution will also have $\cos x$ as a solution, giving us five independent solutions. The other is to explicitly check that a fourth order equation of the type I described in the previous paragraph would not hold, by computing

$$\frac{d^4}{dx^4}[\sin x]+a_1 \frac{d^3}{dx^3}[\sin x] =\sin x - a_1 \cos x$$

which will never be zero by the independence of $\sin$ and $\cos$. This tells us that the order can be no smaller than $5$. Lastly, we can check that

$$y^{(5)}+y^{(3)}=0$$

does indeed have all of the given solutions, so five is the minimal order.

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Hint. The (real) characteristic polynomial should have as roots $\pm i$ (for $\sin(x)=(e^{ix}-e^{-ix})/(2i)$) and $0$ with multiplicity at least 3 (for $x^3$ and $1$).

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