# How to find solutions of higher-order homogeneous ODEs (combinations of real, repeated, AND imaginary roots)

We previously worked on solving different cases where we had second-order ODEs we could find a general solution for, and they had distinct cases for real distinct, real repeated, or complex roots.

What about a higher-order ODE that has a combination of these kinds of roots together?

Let’s work on some examples of such cases.

Example problem 1: Find the general solution of $$y”’ + 4y”’ + 6y” + 4y’ + y = 0$$

Let $$y = {e^{mx}}$$, then find the derivatives and substitute those into the problem above, and you will get a polynomial you can factor.

${m^4} + 4{m^3} + 6{m^2} + 4m + 1 = 0$

You might not be asked often to do this by hand, if you have software or a scientific calculator you can check the roots, or you can try plugging in numbers to find 0 = 0. You can even do long division to find the roots.

But you will find that this polynomial actually has 4 real roots, all of them being -1. So it’s a root of -1 with multiplicity 4.

${(m + 1)^4} = 0$

This is an example of repeated roots, but more than 2 repeated roots, you just need to substitute them into the general solution and just keep adding higher exponents on x terms.

$y = {c_1}{e^{ – x}} + x{c_2}{e^{ – x}} + {x^2}{c_3}{e^{ – x}} + {x^3}{c_4}{e^{ – x}}$

Example problem 2: Find the general solution of $$y”” + 2y” + y = 0$$

Let $$y = {e^{mx}}$$, then find the derivatives and substitute them in, and you will get a polynomial you can factor.

${m^4} + 2{m^2} + 1 = 0$

${({m^2} + 1)^2} = 0$

There will actually be four imaginary roots, 2 are the imaginary number i, and another 2 are -i. So we have repeating roots and complex roots at the same time.

Complex roots will always come in conjugate pairs in the scope of a ODE course.

The value of the real parts is 0, so e to the power of 0 is just 1. This might not always be the case, you would have to multiply the terms by e to the power of the real part if the value of the real part is not 0.

The general solution will be (and you can switch around the constants anywhere):

$y = {c_1}\cos (x) + x{c_2}\cos (x) + {c_3}\sin (x) + x{c_4}\sin (x)$

Try an example of a second-order ODE having both a homogeneous and particular solution!