Method of Solution

To find the solution of the linear homogeneous differential equation wiith constant coefficients $$${{y}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{2}}{y}''+{a}_{{1}}{y}'+{a}_{{0}}={0}$$$, we assume that the solution has the form $$${y}{\left({x}\right)}={{e}}^{{{r}{x}}}$$$.

Plugging this solution into the equation and noting that $$${{y}}^{{{\left({n}\right)}}}={{r}}^{{n}}{{e}}^{{{r}{x}}}$$$ gives

$$${{r}}^{{n}}{{e}}^{{{r}{x}}}+{a}_{{{n}-{1}}}{{r}}^{{{n}-{1}}}{{e}}^{{{r}{x}}}+\ldots+{a}_{{2}}{{r}}^{{2}}{{e}}^{{{r}{x}}}+{a}_{{1}}{r}{{e}}^{{{r}{x}}}+{a}_{{0}}{{e}}^{{{r}{x}}}={0}$$$, or

$$${{r}}^{{n}}+{a}_{{{n}-{1}}}{{r}}^{{{n}-{1}}}+\ldots+{a}_{{2}}{{r}}^{{2}}+{a}_{{1}}{r}+{a}_{{0}}={0}$$$.

This is called the characteristic equation of the differential equation. Solving it, we will find $$${n}$$$ solutions $$${r}_{{1}},{r}_{{2}},\ldots,{r}_{{n}}$$$; so, the solutions of the homogeneous equation are $$${y}_{{1}}{\left({x}\right)}={{e}}^{{{r}_{{1}}{x}}},\ {y}_{{2}}{\left({x}\right)}={{e}}^{{{r}_{{2}}{x}}},\ \ldots,\ {y}_{{n}}{\left({x}\right)}={{e}}^{{{r}_{{n}}{x}}}$$$.

Depending on the roots of the characteristic equation, there are different general solutions.

Case 1. The roots are distinct and real. In this case, $$${y}_{{1}}{\left({x}\right)}={{e}}^{{{r}_{{1}}{x}}},\ {y}_{{2}}{\left({x}\right)}={{e}}^{{{r}_{{2}}{x}}},\ \ldots,\ {y}_{{n}}{\left({x}\right)}={{e}}^{{{r}_{{n}}{x}}}$$$ are all different and thus linearly independent (exponents with different powers are linearly independent); so, the solution can be written as $$${y}={c}_{{1}}{{e}}^{{{r}_{{1}}{x}}}+{c}_{{2}}{{e}}^{{{r}_{{2}}{x}}}+\ldots+{c}_{{n}}{{e}}^{{{r}_{{n}}{x}}}$$$.

Case 2. The roots are complex. Since the coefficients in the differential equation are assumed to be real, there is a complex root $$${a}+{b}{i}$$$ and the corresponding complex conjugate $$${a}-{b}{i}$$$. For this pair of roots, the part of the solution is given by $$${d}_{{1}}{{e}}^{{{\left({a}+{b}{i}\right)}{x}}}+{d}_{{2}}{{e}}^{{{\left({a}-{b}{i}\right)}{x}}}$$$.

Now, using Euler's formula, we can write that $$${{e}}^{{{\left({a}+{b}{i}\right)}{x}}}={{e}}^{{{a}{x}}}{{e}}^{{{b}{i}{x}}}={{e}}^{{{a}{x}}}{\left({\cos{{\left({b}{x}\right)}}}+{i}{\sin{{\left({b}{x}\right)}}}\right)}$$$ and $$${{e}}^{{{\left({a}-{b}{i}\right)}{x}}}={{e}}^{{{a}{x}}}{{e}}^{{-{b}{i}{x}}}={{e}}^{{{a}{x}}}{\left({\cos{{\left({b}{x}\right)}}}-{i}{\sin{{\left({b}{\left({x}\right)}\right)}}}\right.}$$$.

So, $$${d}_{{1}}{{e}}^{{{\left({a}+{b}{i}\right)}{x}}}+{d}_{{2}}{{e}}^{{{\left({a}-{b}{i}\right)}{x}}}={\left({d}_{{1}}+{d}_{{2}}\right)}{{e}}^{{{a}{x}}}{\cos{{\left({b}{x}\right)}}}+{\left({d}_{{1}}-{d}_{{2}}\right)}{i}{{e}}^{{{a}{x}}}{\sin{{\left({b}{x}\right)}}}$$$.

Taking $$${c}_{{1}}={d}_{{1}}+{d}_{{2}}$$$ and $$${c}_{{2}}={\left({d}_{{1}}-{d}_{{2}}\right)}{i}$$$, we can write the solution as $$${c}_{{1}}{{e}}^{{{a}{x}}}{\cos{{\left({b}{x}\right)}}}+{c}_{{2}}{{e}}^{{{a}{x}}}{\sin{{\left({b}{x}\right)}}}$$$.

Case 3. Some roots have the multiplicity $$${p}>{1}$$$ (if the root $$${r}_{{k}}$$$ is of the multiplicity $$${p}$$$, we have that $$${{\left({r}-{r}_{{k}}\right)}}^{{p}}$$$ is a factor of the characteristic equation, but $$${{\left({r}-{r}_{{k}}\right)}}^{{{p}+{1}}}$$$ is not). In this case, some solutions are the same and, therefore, are not linearly independent. Here, the solutions are $$${y}_{{1}}={{e}}^{{{r}_{{k}}{x}}}$$$, $$${y}_{{2}}={x}{{e}}^{{{r}_{{k}}{x}}}$$$, ...,$$${y}_{{p}}={{x}}^{{{p}-{1}}}{{e}}^{{{r}_{{k}}{x}}}$$$.

Now, let's do some more work.

Note that, in general, there will be some roots of the multiplicity more than one, some roots are complex, and some are real and distinct. In this case, we just combine the corresponding solutions.

In theory, it is always possible to factor the characteristic equation, but in practice this can be extremely difficult, especially for differential equations of higher order. In such cases, one often has to use numerical techniques to approximate the solutions.