Recall:
- A Higher-order linear differential equation with constant coefficients of the standard form $y^{(n)} +a_1y^{(n-1)}+ \dots a_{(n-1)} y^{(1)}=0$ is called a homogeneous equation, where $a_1,a_2, \dots, a_{(n-1)} \in \mathbb{R}$ are constants.
- A Higher-order linear differential equation with constant coefficients of the standard form $y^{(n)} +a_1y^{(n-1)}+ \dots a_{(n-1)} y^{(1)}=g(t)$ is called a non-homogeneous equation, where $a_1,a_2, \dots, a_{(n-1)} \in \mathbb{R}$ constants, and $g(t)$ is a function of $t$.
So far, we have covered the solutions of homogeneous equations. This is significant, as we also make use of the solutions of the homogeneous form of a non-homogeneous equation while solving.
Consider a non-homogeneous equation given by
$$ y^{(n)} +a_1y^{(n-1)}+ \dots a_{(n-1)} y^{(1)}=g(t). $$
Let the homogeneous solution of the equation be $y_h$ and the particular solution be $y_p$. We say that the general solution of the equation is given by
$$ y=y(t)=y_h+y_p. $$
Here, it is important to underline the existence of a particular solution. A particular solution represents a solution that inherently satisfies the equation when we plug $y=y_p$. The reason for its existence is basically the term $g(t)$.
The method of undetermined coefficients to solve such equations is as follows:
Consider a non-homogeneous equation given by
$$ y^{(n)} +a_1y^{(n-1)}+ \dots a_{(n-1)} y^{(1)}=g(t). $$
- Convert the equation into the homogeneous form, i.e., $y^{(n)} +a_1y^{(n-1)}+ \dots a_{(n-1)} y^{(1)}=0$. Then, find the homogeneous solution. Call it $y_h$.
- Guess a particular solution, inspecting the explicit expression of $g(t)$ (We will learn about this). Call it $y_p$.
- Construct the final general solution as $y=y_h+y_p$.
An example to gain insight: Consider the higher-order linear non-homogeneous differential equation given by
$$ y^{\prime\prime}-y=1. $$
We wish to find a general solution using the method of undetermined coefficients. Following the strategy mentioned above, consider the homogeneous equation
$$ y^{\prime\prime}-y=0. $$
Write down the characteristic equation by $\lambda^2-1=(\lambda -1)(\lambda +1)=0$ and obtain the roots $\lambda_1=1$ and $\lambda_2=-1$. Therefore, the homogeneous solution is given by
$$ y_h=c_1e^{t} + c_2e^{-t}, $$
where $c_1$ and $c_2$ are arbitrary constants. Now, let us think for a moment. What do we need as a particular solution to get $(particular)^{\prime\prime}-(particular)=1$? Guess that $y_p=A$, where $A$ is a constant. Plugging in, we get
$$ (A)^{\prime\prime}-(A)=1 \implies -A=1 \implies A=-1 $$
So, we get $A=-1$. This implies that the particular solution is given by $y_p=-1$. Hence, the general solution of the equation is given by
$$ y=y(t)=y_h+y_p=c_1e^{t} +c_2e^{-t} -1. $$
Of course, we usually do not try to derive a different guessed particular solution for each differential equation. We have a list where specific guessed particular solutions are noted for corresponding forcing terms; here, the forcing term stands for what we have denoted as $g(t)$. Consider the differential equation
$$ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y^{\prime} + a_0 y = f(t). $$
| Forcing term | Guess for $y_p(t)$ |
|---|---|
| $C$ | $A$ |
| $t^n$ | $A_n t^n + \dots + A_1 t + A_0$ |
| $e^{at}$ | $A e^{at}$ |
| $\sin(bt), \cos(bt)$ | $A\cos(bt) + B\sin(bt)$ |
| $e^{at} t^n$ | $e^{at}(A_n t^n + \dots + A_0)$ |
| $e^{at}\cos(bt), e^{at}\sin(bt)$ | $e^{at}(A\cos(bt) + B\sin(bt))$ |
| $\text{Polynomial}\cdot \sin(bt)$ | $\text{Polynomial}\cdot \sin(bt)$ |
| $\text{Polynomial}\cdot e^{at}$ | $\text{Polynomial}\cdot e^{at}$ |
Resonance Rule
Let $y_h$ be the solution of
$$ a_n y^{(n)} + \dots + a_0 y = 0. $$
If the guessed form of $y_p$ already appears in $y_h$, multiply the entire guess by $t$. If the corresponding characteristic root has multiplicity $m$, multiply the guess by $t^m$.
$$ y(t) = y_h(t) + y_p(t). $$
Example (1)
Find the general solution to the non-homogeneous differential equation given as
$$ y^{\prime\prime}+5y^{\prime}+4y=2t+3. $$
Solution: Following the method, we first construct the corresponding homogeneous equation. We get
$$ y^{\prime\prime}+5y^{\prime}+4y=0. $$
Now, the characteristic equation is given as $\lambda^2+5\lambda+4=(\lambda+1)(\lambda+4)=0$. The roots are $\lambda_1=-1$ and $\lambda_2=-4$. Hence, the homogeneous solution is given by
$$ y_h=c_1e^{-t} + c_2e^{-4t}, $$
where $c_1,c_2$ are arbitrary constants. Secondly, we wish to guess a particular solution. Let us inspect the forcing term, i.e., $2t+3$. We see that the forcing term involves a polynomial of degree 1. So, by the table above, we guess the particular solution of the differential equation as $y_p=At+B$, where $A$ and $B$ are constants.
We want to plug in the particular solution to the equation in order to find the constants $A$ and $B$. Notice that
$$ \begin{aligned} y_p^{\prime} &= A \\ y_p^{\prime\prime} &= 0 \end{aligned} $$
and so, plugging in,
$$ \begin{aligned} 0 + 5(A) + 4(At+B) &= 2t+3 \\ 5A + 4At + 4B &= 2t +3 \\ &\begin{cases} 4A=2 \\ 5A+4B=3 \end{cases} \end{aligned} $$
Solving the equations, we obtain $A=1/2$ and $B=1/8$. Therefore, the particular solution is $y_p=\frac{t}{2} +\frac{1}{8}$. Hence, finally, the general solution of the differential equation is given by
$$ y=y_h+y_p=c_1e^{-t} + c_2e^{-4t} +\frac{t}{2} +\frac{1}{8}. $$ $\square$
Example (2)
Solve the initial value problem
$$ y^{\prime\prime}-4y=8e^{2t}, \quad y(0)=y^{\prime}(0)=1. $$
Solution: We first construct the corresponding homogeneous equation as
$$ y^{\prime\prime}-4y=0. $$
The characteristic equation is given by $\lambda^2-4=(\lambda -2)(\lambda+2)=0$. The roots are clearly $\lambda_1=2$ and $\lambda_2=-2$. Therefore, the homogeneous solution is given by $y_h=c_1e^{2t}+c_2e^{-2t}$, where $c_1,c_2$ are arbitrary constants.
Now, we wish to guess a particular solution. Consider $Ae^{2t}$. However, we observe that the homogeneous solution already includes a term in which $e^{2t}$ occurs, namely $c_1e^{2t}$. Therefore, we must multiply $Ae^{2t}$ by $t$ and guess the particular solution as $y_p=t\left(Ae^{2t}\right)$, where $A$ is a constant.
Now, we get
$$ \begin{aligned} y_p^{\prime} &=Ae^{2t} + t\left(2Ae^{2t}\right) \\ y_p^{\prime\prime} &=2Ae^{2t} + 2Ae^{2t} + t\left(4Ae^{2t}\right)=\left(4Ae^{2t}\right)(t+1) \end{aligned} $$
and then, plugging in the particular solution,
$$ \begin{aligned} \left(4Ae^{2t}\right)(t+1) - 4\left(tAe^{2t}\right) &=4Ae^{2t} \\ &=8e^{2t}. \end{aligned} $$
Hence, $A=2$, and $y_p=2te^{2t}$. Finally, the general solution is given by
$$ y(t)=y_h+y_p=c_1e^{2t}+c_2e^{-2t} +2te^{2t}. $$
Now, we will consider the initial conditions $y(0)=y^{\prime}(0)=1$ and find the solution of the IVP by determining $c_1$ and $c_2$. Consider
$$ \begin{aligned} y(0) &= c_1 + c_2 = 1 \\ y^{\prime}(0) &=2c_1-2c_2 +2=1 \implies y^{\prime}(0)=c_1-c_2=-\frac{1}{2}. \end{aligned} $$
Solving, we obtain $c_1=\frac{1}{4}$ and $c_2=\frac{3}{4}$. Hence, the solution of the IVP is given by
$$ y=\frac{1}{4}e^{2t}+\frac{3}{4}e^{-2t} +2te^{2t}. $$ $\square$