Characteristic Equations
One method of solving linear homogeneous differential equations with constant coefficients is by finding the corresponding "characteristic equation."
For a differential equation written as `\displaystyle\sum_{k=0}^na_ny^{(n)}=0` with constant coefficients `a_n`,
the characteristic equation is given by `p(z)=\displaystyle\sum_{k=0}^na_nz^{n}`.
Assume that `a_0=1` for simplicity. If `a_0\ne1`, then divide the entire equation by `a_0` to obtain a new one which satisfies the assumption.
Using the differential operator `D=\mathrm d/\mathrm dx`, we can see that `p(D)=D^n+a_1D^{n-1}+\ldots+a_nI` where `I` is the identity operator.
With this in mind, we can rewrite the differential equation as `p(D)(y)=0`. Hence, finding a function which "kills" `p(D)` (sends it to zero) is an equivalent task to solving the differential equation.
Using the "ansatz" (educated guess) of `y=e^{rt}`, we can try evaluating `p(D)(e^{rt})` below.
`p(D)(e^{rt})=D^n(e^{rt})+a_1D^{n-1}(e^{rt})+\ldots+a_n(e^{rt})=r^n(e^{rt})+a_1r^{n-1}(e^{rt})+\ldots+a_n(e^{rt})=p(r)\cdot e^{rt}\implies\boxed{p(D)(e^{rt})=p(r)e^{rt}}`
This finding is of tremendous importance, and for this reason it is sometimes called the "key identity."
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(Note that in a sense, one can think of this as stating that `e^{rt}` is an eigenvector of the linear mapping `p(D)` from the set of infinitely differentiable functions to itself.)
Recall that the task at hand was finding a function which kills `p(D)`. Using the key identity, we know that this is accomplished with the function `y=e^{rt}` whenever `r` is a root of `p`, the characteristic polynomial.
This is a great breakthrough, but there are now some different cases we must consider, depending on the distinctness of the roots and whether they are real or complex.
Distinct Roots
When all roots of the characteristic equation are distinct, we can prove easily that this method generates a fundamental set of solutions.
Let `e^{r_1t}, e^{r_2t},\ldots,e^{r_nt}` be solutions found for an `n`
th-order differential equation using the key identity.
It can be shown that the linear independence of these exponentials
follows directly from
the fact that
eigenvectors which correspond to different eigenvalues are linearly independent.
In a sense, one can think of this as stating that `e^{rt}` is an eigenvector of the linear `D` operator with distinct eigenvalue `r`. The fact that different eigenvalues implies
linear independence is intuitively obvious. Take eigenvectors `\vec v_1` and `\vec v_2` with eigenvalues `a` and `b`, respectively. If `\vec v_1` and `\vec v_2` were scalar multiples of one another,
then applying the transformation to the shared unit vector `\vec u=\dfrac{\vec v_1}{\|\vec v_1\|}=\dfrac{\vec v_2}{\|\vec v_2\|}` should result in `T(u)=a\vec u=b\vec u`.
From this, we immediately gather that `a=b`, implied from the fact that the two eigenvectors shared a direction. Thus, using the contrapositive, we find that `a\ne b` implies `v_1` and `v_2` are linearly independent.
Since there are `n` linearly independent solutions to an `n`
th-order differential equation, they must form a fundamental set.
Therefore, each distinct root of the characteristic equation corresponds to a term in the general solution to the differential equation. For roots `\alpha_0`, `\alpha_1`, `\alpha_2`, etc.,
the general solution is `\displaystyle y=c_0e^{\alpha_0x}+c_1e^{\alpha_1x}+c_2e^{\alpha_2x}+\cdots` for constant coefficients `c_n`.
Repeated Roots
Using the fundamental theorem of algebra, let `p(z)=(z-r_1)^{m_1}(z-r_2)^{m_2}\cdots(z-r_i)^{m_i}\cdots(z-r_k)^{m_k}` where we have multiplicities `m\in\mathbb N_+` and at least one `m` is greater than `1`. Without loss of generality, let one such `m` be `m_i`.
Rewrite the function as `p(z)=a(z)(z-r_i)^{m_i}` for an arbitrary polynomial `a(z)` created by grouping together the other factors. Differentiating, we obtain
`p'(z)=a'(z)(z-r_i)^{m_i}+m_ia(z)(z-r_i)^{m_i-1}=\left(a'(z)(z-r_i)+m_ia(z)\right)\cdot(z-r_i)^{m_i-1}`. We can conclude that, since `m_i-1>0`, we know `r_i` is still a root of `p'(z)`.
This fact will allow us to generate another solution to the differential equation based on the original. Let `L=p(D)` for clarity, so that we can write `L(e^{rt})=p(r)e^{rt}`. Taking the
partial derivative with respect to `r` on both sides, we find that `L(te^{rt})=p'(r)e^{rt}+tp(r)e^{rt}`. Hence, any repeated root `r` of the characteristic equation `p` will kill both terms
on the right-hand side, and it is proven that `te^{rt}` is also a solution to the differential equation. In general, using this technique with a root of multiplicity `m` can generate `m` solutions
in the form `e^{rt},te^{rt},\ldots,t^{m-1}e^{rt}`, and doing this for each root will result in a fundamental set of solutions. In other words, if there are repeated roots, then there is not an arbitrary coefficient, but rather an arbitrary polynomial in `t`.
For instance, the characteristic equation `p(r)=(r-2)^3(r-5)` provides us with the general solution `y=c_0t^2e^{2t}+c_1te^{2t}+c_2e^{2t}+c_3e^{5t}`, which can be written more simply as `y=e^{2t}\left(c_0t^2+c_1t+c_2\right)+c_3e^{5t}`.
Complex Roots
When some or all of the roots of the characteristic equation are complex, we are primarily interested in using these results to recover real-valued solutions to the differential equation. In fact, assuming that
the coefficients of the differential equation are all real, we can make a number of developments toward this end. Since all coefficients of the characteristic equation are real, we know that its complex roots
come in conjugate pairs, `r=a+bi` and `\overline r=a-bi`.
This fact about the complex conjugate pairs may not be immediately obvious. Note that what we are trying to prove is that `p(z)=0\implies p(\overline z)=0`. Start with `p(z)=z^n+a_1z^{n-1}+\ldots+a_n=0` and
take the conjugate of both sides, giving `\overline{z^n+a_1z^{n-1}+\ldots+a_n}=0`. By the properties `\overline{x+y}=\overline x + \overline y` and `\overline{x\cdot y}=\overline x\cdot\overline y`, we can see that `{\overline z}^n+a_1{\overline z}^{n-1}+\ldots+a_n=p(\overline z)=0`,
as we set about to prove.
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We can then state that, by Euler's formula, `e^{\overline rt}=\overline{e^{rt}}`, so if `y=e^{rt}` is a solution, then so is `\overline y`. Using the facts that `\mathrm{Re}(z)=\dfrac{z+\overline z}2` and `\mathrm{Im}(z)=\dfrac{z-\overline z}{2i}`,
we see that `\mathrm{Re}(y)=e^{at}\cos(bt)` and `\mathrm{Im}(y)=e^{at}\sin(bt)` are each linear combinations of the solutions `y` and `\overline y`, and it can be shown
that this pair will also be linearly independent for `b\ne0`.
The linear independence of these new solutions can be proven quite easily. Assume the contrary, so that there exist some constants `c_1` and `c_2` such that `c_1e^{at}\cos(bt)+c_2e^{at}\sin(bt)=0` and `c_1` and `c_2` are not both zero.
Since `e^{at}\ne0`, we can divide by `e^{at}` to obtain `c_1\cos(bt)+c_2\sin(bt)=0`. When `t=0`, we have `c_1=0`, so `c_2\sin(bt)=0`. Take `b\ne0`, so that at `t=\dfrac{\pi}{2b}` we have `c_2=0` and the
two constants `c_1` and `c_2` are both zero, so the two solutions are linearly independent. On the other hand, if `b=0`, then `\mathrm{Im}(y)` simply equals zero, and the two are not linearly independent.
However, this corresponds to the case where `r` has no complex part, so this case can effectively be ignored.
Summary
In general, these different cases can be combined into one rule to follow for all roots of the characteristic equation, regardless of the multiplicity or whether it is real or complex.
If `r=a+bi\in\mathbb C` is a root with multiplicity `m`, then the differential equation has solutions `e^{at}\cos(bt),te^{at}\cos(bt),\ldots,t^{m-1}e^{at}\cos(bt)` and `e^{at}\sin(bt),te^{at}\sin(bt),\ldots,t^{m-1}e^{at}\sin(bt)`.
This results in `m` linearly independent solutions for each real root (since the `\sin(bt)` terms are all zero and contribute nothing) and `2m` linearly independent solutions for each pair of complex roots
of the characteristic equation. (Note that this implies that a linear homogeneous differential equation with constant coefficients of degree `d` will have exactly `d` linearly independent solutions.)
TODO: Variation of parameters, method of undetermined coefficients, fundamental/natural solutions
Information heavily drawn from the math department of the University of Maryland at College Park.