As with 2^nd order differential equations we can't determine a non-homogeneous differential equation unless we can initially solve the homogeneous differential equation. We'll also require to restrict ourselves down to constant coefficient differential equations since solving non-constant coefficient differential equations is quite hard and thus we won't be discussing them now. Similarly, we'll only be searching for linear differential equations.

Therefore, let's start off with the subsequent differential equation,

a_n y⁽ⁿ⁾ +a_n y^(n-1)+     .....................      +a₁ y' +  a₀ y = 0

 Here, suppose that solutions to this differential equation will be in the form y(t) = e^rt and plug that in the differential equation and along with a little simplification we find,

e^rt = (a_n rⁿ +a_n r^(n-1)+     .....................      +a₁ r +  a₀) = 0

And therefore in order for this to be zero we will require that,

a_n rⁿ +a_n r^(n-1)+     .....................      +a₁ r +  a₀ = 0

This is termed as the characteristic polynomial/equation and its roots/solutions will provides us the answers to the differential equation. We know that, containing repeated roots, an n^th degree polynomial that we have now will have n roots. Therefore, we require going through all the possibilities which we've found for roots here.

It is where we start to notice differences in how we deal along with n^th order differential equations versus 2^nd order differential equations. There are even the three major cases as: real distinct roots, complex roots and repeated roots (though these can here also be repeated as we'll notice). In 2^nd order differential equations all differential equations could only include one of these cases. However, now, this will not essentially be the case. We could very simply have differential equations which include each of these cases.

For illustration assume that we have an 9^th order differential equation. The whole list of roots could contain 3 roots that only occur once in the list that is real distinct roots, a root with multiplicity 4 that is happens 4 times in the list and a set of complex conjugate roots (recall this since the coefficients are all real complex roots will always happen in conjugate pairs).

Therefore, for each nth order differential equation we'll require to form a set of n linearly independent functions (that is a fundamental set of solutions) in order to find a general solution. In the work which follows we'll discuss the solutions that we find from all cases but we will leave this to you to verify that while we put everything together to form a general solution which we do indeed find a fundamental set of solutions.  Recall this that we need to verify this Wronskian is not zero.

Therefore, let's get started with the work now. Let's start off through assuming that in the list of roots of the characteristic equation we contain, r₁, r₂,  ....   r_k and they only happen once in the list. The solution from all of these will be after that,

Here is nothing really new now for real distinct roots.

Here we take a look at repeated roots. The result this time is a natural extension of the work we noticed in the 2^nd order case. Let's assume that r is a root of multiplicity k (that is r happens k times in the list of roots). We will after that gets the following k solutions to the differential equation as,

e^rt, t e^rt,    ........., t^k-1 e^rt

Therefore, for repeated roots we only add in a t for all of the solutions past the first one till we have a sum of k solutions. Again, we will leave this to you to calculate the Wronskian to verify that such are actually a set of linearly independent solutions.

At last we require dealing with complex roots. The biggest matter now is that we can here have repeated complex roots for 4^th order or higher differential equations. We'll start off through assuming that r = l + mi arises only once in the list of roots. For this case we'll find the standard two solutions as,

e^lt cos(mt)        e^lt sin(mt)

Here let's assume that r = l + mi has a multiplicity of k (that is they arise k times in the list of roots). In that case we can utilize the work from the repeated roots above to find the following set of 2k complex-valued solutions,

e^{(l + mi)t},  t e^{(l + mi)t},    ........., t^k-1 e^{(l + mi)t}

e^{(l - mi)t},  t e^{(l - mi)t},    ........., t^k-1 e^{(l - mi)t}

The problem now obviously is that we really need real-valued solutions. Therefore, recall this in the case where they occurred one time all we had to do was use Euler's formulas on the first one and after that take the real and imaginary part to find two real valued solutions.  We'll do similar thing here and use Euler's formula on the first set of complex-valued solutions given above, split all one in its real and imaginary parts to get there at the following set of 2k real-valued solutions.

e^lt cos(mt),       e^lt sin(mt),        t e^lt cos(mt),     t e^lt sin(mt),     ............,

                        t^k-1 e^lt cos(mt),             t^k-1 e^lt sin(mt),

Once again we'll leave this to you to verify that these do actually form a fundamental set of solutions.

Before we work a couple of quick illustrations now we must point out that the characteristic polynomial is this time going to be at least a 3^rd degree polynomial and determining the roots of these by hand is frequently a very hard and time consuming process and in several cases if the roots are not rational (that is in the form p) this can be almost not possible to get them all by hand. In practice though, we usually use several form of computation aid as Maple or Mathematica to get all the roots.