Number of solutions to the order of the equation

[Trying2Learn]

TL;DR Summary

Why do we expect n solutions to an n-th order differential equation

For a linear nth order differential equation with constant coefficients, the general solution can be expressed as a linear combination of n linearly independent solutions.

Fine...

By finding n linearly independent solutions, we are essentially covering all the necessary components to form the general solution. These solutions form a basis for the solution space, enabling us to express any solution to the differential equation as a linear combination of them. This approach ensures that you capture all the possible behaviors and variations allowed by the equation.

Fine... I get that.

For example, if it is a second order, we have two remaining "unknowns": the first derivative and the zero derivative (the equation itself).

Thus, it seems "justified" in searching for two.

However, consider the case, in a second order (spring, mass, damping), when the roots of the characteristic, repeat. We find a second solution by multiplying the first, by say, time. No problem.

However, my question is that I do NOT see this "search" for a second solution, as motivated by finding expressions for the first or zeroth order.
The process just seems whimsical.

I suppose I am stuck between two issues: 1) why must there be as many solutions as order of the equation and 2) is the search for other solutions grounded in the exploration of "n" solutions (all the the n-1, n-2, n-3, 0) terms?

In fact, to go further, when we do get the quadratic characteristic equation for an oscillator, the act of claiming two solutions seems serendipitous, rather than grounded in a search for a solution for position, and a second for velocity.

I am not sure I am asking this the right way.

Maybe I should be asking: why do the n solutions form a "basis" and how do we know that basis is of the same order of the differential equation, and why do such searches often seem infused with "guess work?"

 

[pasmith]

There is no guesswork.

A constant coefficient linear ODE of order $n$ can be written as a polynomial in the derivative operator $D$, $P(D)x = f$. This can be factorised over the complex numbers into a product of linear factors $$(D - a_n)(D - a_{n-1}) \cdots (D - a_1)x = f$$ where the $a_i$ are not necessarily distinct. This can be written as an $n$-dimensional first-order system for $(x, y_1, \dots, y_{n-1})$, $$\begin{split} (D - a_1)x &= y_1 \\ (D - a_2)y_1 &= y_2 \\ &\vdots \\ (D - a_n)y_{n-1} &= f. \end{split}$$ Starting with the last, we can solve each of these in turn by an integrating factor: $$\begin{split} y_{n-1}(t) &= e^{a_nt}\int^t e^{-a_ns}f(s)\,ds \\ y_{n-2}(t) &= e^{a_{n-1}t}\int^t e^{-a_{n-1}s}y_{n-1}(s)\,ds \\ &\vdots \\ x(t) &= e^{a_1t} \int^t e^{-a_1 s} y_1(s)\,ds.\end{split}$$ It is instructive to try this yourself for the case $$(D - a)^2x = Ae^{at} + Be^{bt}, \quad b \neq a.$$ If the coefficients are real, then the $a_i$ must either be real or they must appear in complex conjugate pairs $p \pm iq$. Here we make use of Euler's formula and write $$e^{(p \pm iq)t} = e^{pt}(\cos qt \pm i \sin qt)$$ noting that for a solution to start real and remain real the coefficients of conjugate functions must be conjugates. This allows us to write the general real-valued solution as a linear combination of sines and cosines. If sines or cosines appear on the right hand side, then we do the reverse and write $$\begin{split} e^{pt}\cos (qt) &= \frac{e^{(p + iq)t} + e^{(p - iq)t}}{2} \\ e^{pt}\sin (qt) &= \frac{e^{(p + iq)t} - e^{(p - iq)t}}{2i}. \end{split}$$
Although this method is guaranteed to work if we can find the roots of $P$, it is not always the easiest method. Rather, the easiest method is to apply pattern recognition borne of experience, which often appears as guesswork to those who have not yet gained such experience.

 

[Erland]

So, let us look at two specific examples and use pasmith's method:

Example 1.
$$y''-5y'+6y=0$$

The characterstic equation $r^2-5r+6=0$ has two distinct roots: $r_1=2$ and $r_2=3$.
Therefore, the characteristic polynomial can be factorized $(r-2)(r-3)$.
Now, we can calculate with differential operators in the same way as with polynomials, så our differential eqaution can be written:
$$(D^2-5D+6)y=0$$ so
$$(D-2)(D-3)y=0$$
If we put $u=(D-3)y$, we get
$$(D-2)u=0$$ or
$$u'-2u=0$$
We solve this multiplying by the integrating factor $e^{-2x}$. Then,
$$u'e^{-2x}-2ue^{-2x}=0$$ and, by using the product rule backwards:
$$(ue^{-2x})'=0$$
Integrating gives:
$$ue^{-2x}=C_0$$ and multiplying by $e^{2x}$
$$u=C_0\,e^{2x}$$
But $u=(D-3)y=y'-3y$, so
$$y'-3y=C_0\,e^{2x}$$
Now, we have the integrating factor $e^{-3x}$, so
$$y'e^{-3x}-3ye^{-3x}=C_0\,e^{2x}e^{-3x}$$
$$(ye^{-3x})'=C_0\,e^{-x}$$
Integrating:
$$ye^{-3x}=-C_0\,e^{-x}+D$$ or, with $C=-C_0$,
$$ye^{-3x}=Ce^{-x}+D$$ and, multiplying by $e^{3x}$,
$$y=Ce^{2x}+De^{3x}.$$
So, we got two constants because we integrated twice. and the solution space has dimension 2, with $\{e^{2x},e^{3x}\}$ as a base. We can easily see why the coefficients of the exponents must be the roots of the charateristic equation.

Now, let us instead consider:

Example 2
$$y''-6y'+9y=0$$

The characterstic equation $r^2-6r+9=0$ now has a double root: $r=3$.
Therefore, the characteristic polynomial can be factorized $(r-3)^2$.
Writing this with differential operators gives:
$$(D^2-6D+9)y=0$$ or
$$(D-3)^2y=0$$
This time, we put $u=(D-3)y$, and get
$$(D-3)u=0$$ or
$$u'-3u=0$$
Then, proceeding as in the previous example, we obtain
$$u=Ce^{3x}$$
But this time $u=(D-3)y=y'-3y$, so
$$y'-3y=Ce^{3x}$$
Again, we have the integrating factor $e^{-3x}$, so
$$y'e^{-3x}-3ye^{-3x}=Ce^{3x}e^{-3x}$$
$$(ye^{-3x})'=C.$$
This time, the right hand side is a constant, not an exponential function (with nonzero exponent). We can see that the reason for that is that the characteristic equation has a double root. Integration therefore gives a different form this time:
Integrating:
$$ye^{-3x}=Cx+D$$ and, multiplying by $e^{3x}$
$$y=Cxe^{3x}+De^{3x}$$
or
$$y=(Cx+D)e^{3x}.$$
So, we got two constants here too. and the solution space has dimension 2 here too, with $\{xe^{3x},e^{3x}\}$ as a base. This different form of the solution compared to the previous example is caused by the double root of the characteristic equation.

In general, a linear homogeneous differential equation of degree $n$ with constant coefficients has a general solution of the form
$$y=P_1(x)e^{r_1x}+P_2(x)e^{r_2x}+\dots+P_k(x)e^{r_kx}$$ where $\{r_1,r_2,\dots,r_k\}$ are the distinct roots of the characteristic equation, and $P_1(x), P_2(x),\dots, P_k(x)$ are polynomials, each of degree less than the multiplicity of the corresponding root. So, in Example 2, we had a polyomial $Cx+D$ of degree at most 1, since the root had multiplicity 2.