Homogeneous LODEs with constant coefficients

In this section, we are interested in finding solutions to the homogeneous version of Eq. (13.1) i.e. when $f(x)=0$. We start with the simplest case, where the coefficients are constants (i.e. when $P(x)$ and $Q(x)$ are constants). The differential equation we want to solve is,

By Definition 13.1, the general solution is given by

Our objective here is to find the pair of fundamental solutions $\left[y_{1}(x)\right.$ and $\left.y_{2}(x)\right]$ and use them in Eq. (13.5) to construct the general solution to Eq. (13.18).

Auxiliary equation

For LODEs with constant coefficients like Eq. (13.18), it is, it is easy to see that the equation's solution is a function which, when differentiated, gives us constant multiples of the original function. The statement above motivates us to 'guess' a solution to the ODE that takes the form $y=e^{m x}$ where $m$ is unknown. Our objective is to find $m$.

Differentiating $y=e^{m x}$ wrt $x$ to obtain $y^{\prime}$ and $y^{\prime \prime}$, we get $y^{\prime}=m e^{m x}$ and $y^{\prime \prime}=m^{2} e^{m x}$. Substituting the expressions for $y, y^{\prime}$ and $y^{\prime \prime}$ in Eq. (13.18) and simplifying, we obtain:

Equation (13.19) is satisfied if $e^{m x}=0$ or if $m^{2}+p m+q=0$. Since $e^{m x} \neq 0$, we use the following equation to determine $m$,

Equation (13.20) is known as the characteristic or auxiliary equation of the ODE (13.18) which needs to be satisfied if $y=e^{m x}$ is to be a solution to ODE (13.18). The solution to the quadratic equation gives us the roots, $m$ as follows,

Assuming that $p$ and $q$ are real numbers, we consider three cases:

1. Case 1: The characteristic equation has two real and distinct roots.
2. Case 2: The characteristic equation has two complex roots.
3. Case 3: The characteristic equation has one repeated root.

In what follows, we look at each case separately. Before proceeding we note the following:

• When looking for the solution to a LODE with constant coefficients, the problem is reduced from solving an $O D E$ to solving an algebraic equation [this is the characteristic equation given by (13.20)].
• For second order ODEs we are looking for two fundamental solutions, $y_{1}$ and $y_{2}$.
• The solutions take the form $y_{1}=e^{m_{1} x}$ and $y_{2}=e^{m_{2} x}$ where $m_{1}$ and $m_{2}$ denote the roots to the characteristic equation.
• If there exist two roots to the characteristic equation, the general solution is,

Case 1: Roots are real and distinct

If $p^{2}-4 q>0$, then from Eq. (13.21), we obtain two real roots, $m_{1}$ and $m_{2}$.

Fundamental solutions

The next step is to form the fundamental solutions. Recall that the characteristic equation is a result of using the ansatz, $y=e^{m x}$ where $m$ is unknown, in the differential equation. The two roots are then used to form the fundamental solutions. The first solution is $y_{1}=e^{m_{1} x}$ and the second is $y_{2}=e^{m_{2} x}$; of course any constant multiple of $y_{1}$ (say, $c_{1} y_{1}$ ) and any constant multiple of $y_{2}$ (say, $c_{2} y_{2}$ ) are also solutions to the ODE.

General solution

With knowledge of the pair of fundamental solutions, we can construct the general solution by forming a linear combination.

If the roots of the characteristic equation (13.20), $m_{1}$ and $m_{2}$ are real and distinct, then the general solution to (13.18), is given by

Example 13.2 Find the general solution to the following ODE:

Solution First we notice that the ODE is homogeneous with constant coefficients implying that the solutions to the ODE are exponential and the exponents are the roots to the corresponding characteristic equation. The characteristic equation is,

which gives $m_{1}=2$ and $m_{2}=-1$. The roots are real and distinct so from Eq. (13.22), the general solution is,

Case 2: Roots are complex conjugates

If $p^{2}-4 q<0$, then from Eq. (13.21), we obtain two complex roots, $m_{1}$ and $m_{2}$. These two roots are conjugates,

where $a$ and $b$ are real numbers. Note that $a$ may be zero. As in Case $\mathbf{1}$, this gives two linearly independent solutions: $y_{1}=e^{m_{1} x}$ and $y_{2}=e^{m_{2} x}$. Consequently, the linear combination of $y_{1}$ and $y_{2}$ make up the general solution

Of course a natural question that arises is what becomes of the exponential function if the exponents are complex numbers? The answer to that question is given by Euler's formula, defined below (see also Section 7.2 in Chapter 7).

Definition 13.2 Euler's formula states that for any real number $x, e^{i x}$ is given by,

Equation (13.26) gives the form of the complex roots, $m_{1}$ and $m_{2}$. So, for $m_{1}=a+i b$, the exponential function $y_{1}=e^{m_{1} x}$ becomes,

Similarly, for $m_{2}=a-i b$, we have,

From Eqs. (13.28) and (13.29), we can construct the general solution,

However, the general solution given by (13.30) is a complex-valued function; this means that, given a real number $x$, the value of the function $y(x)$ could be complex. Equation (13.30) represents the most general form: it represents all solutions.

The coefficients in ODE (13.18) are assumed to be real and therefore, we are looking for solutions described by real-valued functions. This means that we want to 'filter out' all functions containing coefficients with an imaginary part, keeping only those whose coefficients are real numbers.

We choose to write down the general solution in the following form*

The two real-valued solutions that we seek therefore are given by

Their Wronskian determinant is $W(x)=b e^{2 a x}$ which is never zero; therefore, the two functions are linearly independent and they constitute a fundamental set of solutions which leads to the general solution given by Eq. (13.31).

If the roots of the characteristic equation (13.20), are complex and distinct i.e. $m_{1}=a+i b$ and $m_{2}=a-i b$, then the general solution to (13.18), is given by

• Here we give reasoning behind the choice of Eq. (13.31) [equivalently, Eq. (13.32)], as the general solution for Case 2. Once we write out the general solution using Euler's formula, we have the following form of the general solution to the ODE [this is (13.30) rearranged]

the coefficients $k_{1}$ and $k_{2}$ are arbitrarily chosen and are also complex. For a second order ODE, we have to satisfy 2 (independent) initial conditions which ultimately dictate what the values of $k_{1}$ and $k_{2}$ are. However, in its most general form, Eq. (13.33) has 4 unknown constants. This can be seen by expanding Eq. (13.33), using $k_{1}=a_{1}+b_{1} i$ and $k_{2}=a_{2}+b_{2} i$ (where $a_{1}, b_{1}, a_{2}$ and $b_{2}$ are real and unknown). Doing so gives some terms that are real and some terms that are imaginary. In order for Eq. (13.33) to represent a real-valued function, the constants $k_{1}$ and $k_{2}$ can no longer be arbitrarily chosen; instead, they must be conjugate numbers such that $k_{1}+k_{2}$ and $i\left(k_{1}-k_{2}\right)$ are real constants. This leaves us with $y_{1}=e^{a x} \cos (b x)$ and $y_{2}=e^{a x} \sin (b x)$ as two real, linearly independent solutions. Their sum using real arbitrary coefficients $c_{1}$ and $c_{2}$ is the real solution we seek [this is given by Eq. (13.31)]. Finally note that the constant $c_{1}$ is $k_{1}+k_{2}$ and $c_{2}$ is $i\left(k_{1}-k_{2}\right)$ where $k_{1}$ and $k_{2}$ are conjugate numbers.

Example 13.3 Find the general solution to the following ODE:

Solution Again, the ODE is homogeneous with constant coefficients so we proceed to solving the corresponding characteristic equation,

which gives the complex conjugate roots $m_{1}=-1+4 i$ and $m_{2}=-1-4 i$. Comparing to the general solution (13.32), $a=-1$ (real part) and $b=4$ (imaginary part) leading to the following solution

where $c_{1}, c_{2}$ are arbitrary constants which can be solved for using initial conditions.

Case 3: Repeated root

If $p^{2}-4 q=0$, then from Eq. (13.21), we obtain one (real) root, $m$ which is equal to $-p / 2$. Again, using the ansatz, $y=e^{m x}$ and $m=-p / 2$, we now have one solution given as,

We now have a problem because unlike Case $\mathbf{1}$ and Case 2, where the solution to the characteristic equation gave us two distinct solutions, here, the characteristic equation only gives one. We therefore do not have enough information to construct the general solution to ODE (13.18) for Case 3. In order to be able to write down the general solution for this case, we need to come up with a second solution (which is, of course, linearly independent to the first one).

Using a known solution to find another

We use the known solution, $y_{1}$ given by Eq. (13.37) to find the second solution. We are looking for $y_{2}$, a linearly independent function to $y_{1}$ which satisfies the ODE (13.18). We assume that the second solution takes the following form

where $v(x)$ is an unknown nonconstant function of $x$, and $y_{1}$ is the solution we already have (from the characteristic equation). The motivation behind this is that if two functions are linearly independent, then, as previously stated in Subsec. 13.1.2, they should not be constant multiples of each other:

The fact that $y_{1}$ and $y_{2}$ are not constant multiples of each other is an encouraging first step in determining the second fundamental solution. The objective here is to determine $v(x)$. If we have $v(x)$ then, by using Eq. (13.38), we have $y_{2}(x)$.

Determine the unknown function, $v(x)$

• If $y_{2}$ is a solution to the ODE (13.18) then it must satisfy Eq. (13.18). Our next step therefore is to take Eq. (13.38) and differentiate wrt $x$ (to obtain $y_{2}^{\prime}$ and $y_{2}^{\prime \prime}$ ) and substitute back in the original ODE,

• Take Eqs. (13.38), (13.40), (13.41) and substitute in (13.18),

• Factorise $v, v^{\prime}$ and $v^{\prime \prime}$ in Eq. (13.42),

Now, Eq. (13.43) is made up of three terms which (from left to right) are: the coefficients of $v$ (blue terms), of $v^{\prime}$ (green terms) and of $v^{\prime \prime}$ (red term).

• Let us start with the coefficients of $v$,

If $y_{1}$ satisfies the ODE (13.18), then,

which means that the coefficient of $v$ is 0 and hence the first term on the LHS of (13.43) disappear,

• Moving on to the coefficient of $v^{\prime}$. This is given by

so, the coefficient of $v^{\prime}$ also disappears,

• We are left with $v^{\prime \prime} y_{1}=0$ and, since $y_{1}=e^{-p x / 2} \neq 0$ then, what we are really left with is $v^{\prime \prime}=0$. This is a second order ODE for $v(x)$ which can be easily solved by integrating twice wrt $x$,

for any nonzero $k_{1}$ and any $k_{2}$. Note that the constants $k_{1}$ and $k_{2}$ can be arbitrarily chosen to construct the general solution since we ultimately solve for the constants appearing in the solution through application of initial conditions. As long as we have $k_{1} \neq 0$ then, $v(x)$ will be a nonconstant function of $x$ which, in turn, guarantees that $y_{1}$ and $y_{2}$ are linearly independent.

• The simplest form $v(x)$ can take therefore is $v(x)=x$ where we chose $k_{1}=1$ and $k_{2}=\overline{0 .}$ This implies that the second solution is given by,

Note that for constant coefficient, homogeneous ODEs whose characteristic equation gives a single, repeated root the pair of fundamental solutions is always given by

where $p$ is the coefficient of $y^{\prime}$ when the ODE is written in standard form.

If the characteristic equation (13.20) gives a repeated, real root i.e. $m=-p / 2$, then the general solution to (13.18), is given by

Example 13.4 Find the particular solution to the following ODE:

Solution In this case, initial conditions are provided at $x=0$, which allow us to solve for a particular solution to the IVP. We start off with the characteristic equation

which gives the a repeated root as $m=3$ yielding the first solution as $y_{1}=e^{3 x}$. A linearly independent solution is $y_{2}=v(x)=y_{1}$ where $v(x)=x$. Using Eq. (13.49), the general solution is

Applying the initial condition, $y(0)=0$ yields $c_{1}=0$ and $y^{\prime}(0)=5$ yields $c_{2}=5$. Thus, the particular solution is:

Note that while the general solution is made up of all linearly independent derivatives, the particular solution satisfying the aforementioned initial conditions is only described by $y_{2}$.

Summary

Given a second order linear, homogeneous equation with constant coefficients,

solve its characteristic equation,

The general solution to (13.54), depends on the type of roots obtained in (13.55).

1. When $p^{2}-4 q>0$, the roots $m_{1}$ and $m_{2}$ are real + distinct and the general solution is,

2. When $p^{2}-4 q<0$, the roots are complex + distinct and they take the form $m=a \pm i b$. The general solution is,

3. When $p^{2}-4 q=0$, there is only one real root given by $m=-p / 2$. The general solution is,

Note the subscript $h$ in the general solutions in Eqs. (13.56)-(13.58). This denotes that the functions, $y_{h}$ are solutions to the homogeneous ODE (i.e. when $f(x)=0$ ). The homogeneous solution $y_{h}$ is also known as the complementary function.

Notes

• You do not need to set the ODE in standard form when solving it using the characteristic equation method. If we start off using the following ODE (not in standard form):

then the characteristic equation becomes,

• In Eq. (13.58), the root given as $m=-p / 2$ is a result of having the ODE in standard form. For the more general form given by Eq. (13.59) and the associated characteristic equation (13.60), the characteristic equation solution is given by,

If $\beta^{2}-4 \alpha \gamma=0$ then the only root is $m=\frac{-\beta}{2 \alpha}$.