# The Method of Undetermined Coefficients for Higher Order Nonhomogenous Differential Equations

Recall from The Method of Undetermined Coefficients page that if we had a second order linear nonhomogenous differential equation whose coefficients were constant, that is $a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = g(t)$, then to solve this differential equation, all we need to do is solve the corresponding second order linear homogenous differential equation $a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = 0$, and then find a partial solution by assuming the form of the particular solution. More precisely, if $g(t)$ was a polynomial, exponential, or sine/cosine function (or a combination of these), then we could assume a form for the particular solutions (see the linked page above for more details) and solve for the coefficients of this form to obtain a particular solution.

We will now look at this method for higher order linear nonhomogenous differential equations. Consider the following $n^{\mathrm{th}}$ order linear nonhomogenous differential equation with coefficients $a_0, a_1, …, a_n \in \mathbb{R}$:

(1)

Suppose that $g(t)$ is of a form containing a polynomial, exponential function, or a sine/cosine function (like with when we were dealing with the method of undetermined coefficients for second order linear nonhomogenous differential equations). We can then find a particular solution $Y(t)$ if we extend the assumed forms of combinations of polynomials, exponential functions, or sine/cosine functions, and multiplying by powers of $t$ to ensure our particular solution does not contain part of a solution to the corresponding higher order linear homogenous differential equation. We can then solve for these constants by plugging our assumed form $Y(t)$ into our differential equation above.

Providing a list of possible forms is trivial, so we will instead look at some examples of apply this method.

## Example 1

**Find the general solution to the differential equation $\frac{d^3y}{dt^3} + \frac{d^2y}{dt^2} + \frac{dy}{dt} + y = e^{-t} + 4t$.**

We will need to first solve the corresponding third order linear homogenous differential equation $\frac{d^3y}{dt^3} + \frac{d^2y}{dt^2} + \frac{dy}{dt} + y = 0$. This characteristic equation to this differential equation is:

(2)

We can (by trial and error) see that $r_1 = -1$ is a solution to this characteristic equation. Applying long division with the factor $(r + 1)$ and we have that our characteristic equation can be written as:

(3)

Therefore we can see that $r_2 = i$ and $r_3 = -i$. Therefore the general solution to our third order linear homogenous differential equation is:

(4)

Now we note that $g(t) = e^{-t} + 4t$ has an exponential term and a cosine term, so we expect the form of our particular solution to be $Y(t) = Ae^{-t} …$. We now compute the first, second, and third derivatives of $Y$.