Second-Order Homogeneous Linear ODEs with Constant Coefficients
We examine how the general solution of homogeneous linear ODEs with constant coefficients takes different forms depending on the sign of the discriminant of the characteristic equation.
TL;DR
- Second-order homogeneous linear ODE with constant coefficients: $y^{\prime\prime} + ay^{\prime} + by = 0$
- Characteristic equation: $\lambda^2 + a\lambda + b = 0$
- The general solution can be categorized into three cases based on the sign of the discriminant $a^2 - 4b$ of the characteristic equation, as shown in the table
Case Roots of characteristic equation Basis of ODE solution General solution of ODE I Distinct real roots
$\lambda_1$, $\lambda_2$$e^{\lambda_1 x}$, $e^{\lambda_2 x}$ $y = c_1e^{\lambda_1 x} + c_2e^{\lambda_2 x}$ II Real double root
$\lambda = -\cfrac{1}{2}a$$e^{-ax/2}$, $xe^{-ax/2}$ $y = (c_1 + c_2 x)e^{-ax/2}$ III Complex conjugate roots
$\lambda_1 = -\cfrac{1}{2}a + i\omega$,
$\lambda_2 = -\cfrac{1}{2}a - i\omega$$e^{-ax/2}\cos{\omega x}$
$e^{-ax/2}\sin{\omega x}$$y = e^{-ax/2}(A\cos{\omega x} + B\sin{\omega x})$
Prerequisites
- Bernoulli Equation
- Homogeneous Linear ODEs of Second Order
- Euler’s formula
Characteristic equation
Let’s consider a second-order homogeneous linear ODE with constant coefficients $a$ and $b$:
\[y^{\prime\prime} + ay^{\prime} + by = 0 \label{eqn:ode_with_constant_coefficients}\tag{1}\]This type of equation is important in applications involving mechanical and electrical oscillations.
Previously, in the Bernoulli Equation, we found the general solution for the logistic equation. According to that, the solution to a first-order linear ODE with constant coefficient $k$:
\[y^\prime + ky = 0\]is the exponential function $y = ce^{-kx}$. (This is the case where $A=-k$ and $B=0$ in equation (4) of that post)
Therefore, for a similar form of equation ($\ref{eqn:ode_with_constant_coefficients}$), we can first try a solution of the form:
\[y=e^{\lambda x}\label{eqn:general_sol}\tag{2}\]Of course, this is just a guess, and there’s no guarantee that the general solution will actually take this form. However, as long as we can find two linearly independent solutions, we can obtain the general solution using the superposition principle, as we saw in Second-Order Homogeneous Linear ODEs.
As we’ll see shortly, there are cases where we need to find solutions in other forms.
Substituting equation ($\ref{eqn:general_sol}$) and its derivatives
\[y^\prime = \lambda e^{\lambda x}, \quad y^{\prime\prime} = \lambda^2 e^{\lambda x}\]into equation ($\ref{eqn:ode_with_constant_coefficients}$), we get
\[(\lambda^2 + a\lambda + b)e^{\lambda x} = 0\]Therefore, if $\lambda$ is a solution to the characteristic equation
\[\lambda^2 + a\lambda + b = 0 \label{eqn:characteristic_eqn}\tag{3}\]then the exponential function ($\ref{eqn:general_sol}$) is a solution to the ODE ($\ref{eqn:ode_with_constant_coefficients}$). Solving the quadratic equation ($\ref{eqn:characteristic_eqn}$), we get
\[\begin{align*} \lambda_1 &= \frac{1}{2}\left(-a + \sqrt{a^2 - 4b}\right), \\ \lambda_2 &= \frac{1}{2}\left(-a + \sqrt{a^2 + 4b}\right) \end{align*}\label{eqn:lambdas}\tag{4}\]From this, we can see that the two functions
\[y_1 = e^{\lambda_1 x}, \quad y_2 = e^{\lambda_2 x} \tag{5}\]are solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$).
Now, we can divide the cases into three categories based on the sign of the discriminant $a^2 - 4b$ of the characteristic equation ($\ref{eqn:characteristic_eqn}$):
- $a^2 - 4b > 0$: Two distinct real roots
- $a^2 - 4b = 0$: Real double root
- $a^2 - 4b < 0$: Complex conjugate roots
Forms of general solution based on the sign of the discriminant of the characteristic equation
I. Two distinct real roots $\lambda_1$ and $\lambda_2$
In this case, the basis of solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$) on any interval is
\[y_1 = e^{\lambda_1 x}, \quad y_2 = e^{\lambda_2 x}\]and the corresponding general solution is
\[y = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} \label{eqn:general_sol_1}\tag{6}\]II. Real double root $\lambda = -\cfrac{a}{2}$
When $a^2 - 4b = 0$, the quadratic equation ($\ref{eqn:characteristic_eqn}$) yields only one solution $\lambda = \lambda_1 = \lambda_2 = -\cfrac{a}{2}$, and thus we can only obtain one solution of the form $y = e^{\lambda x}$:
\[y_1 = e^{-(a/2)x}\]To obtain a basis, we need to find a second solution $y_2$ that is independent of $y_1$.
In this situation, we can use the reduction of order method that we learned earlier. Let’s assume the second solution we’re looking for is of the form $y_2=uy_1$, and substitute
\[\begin{align*} y_2 &= uy_1, \\ y_2^{\prime} &= u^{\prime}y_1 + uy_1^{\prime}, \\ y_2^{\prime\prime} &= u^{\prime\prime}y_1 + 2u^{\prime}y_1^{\prime} + uy_1^{\prime\prime} \end{align*}\]into equation ($\ref{eqn:ode_with_constant_coefficients}$). We get
\[(u^{\prime\prime}y_1 + 2u^\prime y_1^\prime + uy_1^{\prime\prime}) + a(u^\prime y_1 + uy_1^\prime) + buy_1 = 0\]Grouping terms with $u^{\prime\prime}$, $u^\prime$, and $u$, we get
\[y_1u^{\prime\prime} + (2y_1^\prime + ay_1)u^\prime + (y_1^{\prime\prime} + ay_1^\prime + by_1)u = 0\]Here, since $y_1$ is a solution to equation ($\ref{eqn:ode_with_constant_coefficients}$), the expression in the last parenthesis is 0, and
\[2y_1^\prime = -ae^{-ax/2} = -ay_1\]so the expression in the first parenthesis is also 0. Therefore, only $u^{\prime\prime}y_1 = 0$ remains, from which we get $u^{\prime\prime}=0$. Integrating twice, we get $u = c_1x + c_2$, and since the integration constants $c_1$ and $c_2$ can be any values, we can simply choose $c_1=1$ and $c_2=0$ to get $u=x$. Then $y_2 = uy_1 = xy_1$, and since $y_1$ and $y_2$ are linearly independent, they form a basis. Therefore, when the characteristic equation ($\ref{eqn:characteristic_eqn}$) has a double root, the basis of solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$) on any interval is
\[e^{-ax/2}, \quad xe^{-ax/2}\]and the corresponding general solution is
\[y = (c_1 + c_2x)e^{-ax/2} \label{eqn:general_sol_2}\tag{7}\]III. Complex conjugate roots $-\cfrac{1}{2}a + i\omega$ and $-\cfrac{1}{2}a - i\omega$
In this case, $a^2 - 4b < 0$ and $\sqrt{-1} = i$, so from equation ($\ref{eqn:lambdas}$) we have
\[\cfrac{1}{2}\sqrt{a^2 - 4b} = \cfrac{1}{2}\sqrt{-(4b - a^2)} = \sqrt{-(b-\frac{1}{4}a^2)} = i\sqrt{b - \frac{1}{4}a^2}\]Let’s define the real number $\sqrt{b-\cfrac{1}{4}a^2} = \omega$.
With $\omega$ defined as above, the solutions to the characteristic equation ($\ref{eqn:characteristic_eqn}$) are complex conjugate roots $\lambda = -\cfrac{1}{2}a \pm i\omega$, and the corresponding two complex solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$) are
\[\begin{align*} e^{\lambda_1 x} &= e^{-(a/2)x + i\omega x}, \\ e^{\lambda_2 x} &= e^{-(a/2)x - i\omega x} \end{align*}\]However, in this case too, we can obtain a basis of real-valued solutions as follows.
Using Euler’s formula
\[e^{it} = \cos t + i\sin t \label{eqn:euler_formula}\tag{8}\]and the equation obtained by substituting $-t$ for $t$ in the above equation
\[e^{-it} = \cos t - i\sin t\]Adding and subtracting these two equations, we get:
\[\begin{align*} \cos t &= \frac{1}{2}(e^{it} + e^{-it}), \\ \sin t &= \frac{1}{2i}(e^{it} - e^{-it}). \end{align*} \label{eqn:cos_and_sin}\tag{9}\]The complex exponential function $e^z$ of a complex variable $z = r + it$ with real part $r$ and imaginary part $it$ can be defined using the real functions $e^r$, $\cos t$, and $\sin t$ as follows:
\[e^z = e^{r + it} = e^r e^{it} = e^r(\cos t + \sin t) \label{eqn:complex_exp}\tag{10}\]If we let $r=-\cfrac{1}{2}ax$ and $t=\omega x$, we can write:
\[\begin{align*} e^{\lambda_1 x} &= e^{-(a/2)x + i\omega x} = e^{-(a/2)x}(\cos{\omega x} + i\sin{\omega x}) \\ e^{\lambda_2 x} &= e^{-(a/2)x - i\omega x} = e^{-(a/2)x}(\cos{\omega x} - i\sin{\omega x}) \end{align*}\]By the superposition principle, the sum and constant multiples of these complex solutions are also solutions. Therefore, by adding these two equations and multiplying both sides by $\cfrac{1}{2}$, we can obtain the first real solution $y_1$ as follows:
\[y_1 = e^{-(a/2)x} \cos{\omega x}. \label{eqn:basis_1}\tag{11}\]Similarly, by subtracting the second equation from the first and multiplying both sides by $\cfrac{1}{2i}$, we can obtain the second real solution $y_2$:
\[y_2 = e^{-(a/2)x} \sin{\omega x}. \label{eqn:basis_2}\tag{12}\]Since $\cfrac{y_1}{y_2} = \cot{\omega x}$ and this is not a constant, $y_1$ and $y_2$ are linearly independent on all intervals and thus form a basis for the real solutions of equation ($\ref{eqn:ode_with_constant_coefficients}$). From this, we obtain the general solution
\[y = e^{-ax/2}(A\cos{\omega x} + B\sin{\omega x}) \quad \text{(}A,\, B\text{ are arbitrary constants)} \label{eqn:general_sol_3}\tag{13}\]