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 takes three different forms depending on the sign of the discriminant $a^2 - 4b$, as shown in the table
Case Roots of characteristic equation Basis for ODE solutions General solution 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 has important applications in mechanical and electrical oscillations.
Previously in Bernoulli Equation, we found the general solution to the logistic equation, and according to that, the solution to a first-order linear differential equation with constant coefficient $k$:
\[y^\prime + ky = 0\]is the exponential function $y = ce^{-kx}$. (This is the case where $A=-k$, $B=0$ 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 merely 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 determine 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 different 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 differential equation ($\ref{eqn:ode_with_constant_coefficients}$). 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, the two functions:
\[y_1 = e^{\lambda_1 x}, \quad y_2 = e^{\lambda_2 x} \tag{5}\]are solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$).
The terms characteristic equation and auxiliary equation are often used interchangeably, and they mean exactly the same thing. Either term is acceptable.
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 Discriminant
I. Two Distinct Real Roots $\lambda_1$ and $\lambda_2$
In this case, the basis for 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 ($\ref{eqn:characteristic_eqn}$) has only one solution $\lambda = \lambda_1 = \lambda_2 = -\cfrac{a}{2}$, and therefore we can only obtain one solution of the form $y = e^{\lambda x}$:
\[y_1 = e^{-(a/2)x}\]To find a basis, we need to discover a second solution $y_2$ that is independent of $y_1$.
In this situation, we can use the reduction of order technique we learned earlier. Let’s set the second solution as $y_2=uy_1$ and find:
\[\begin{align*} y_2 &= uy_1, \\ y_2^{\prime} &= u^{\prime}y_1 + uy_1^{\prime}, \\ y_2^{\prime\prime} &= u^{\prime\prime}y_1 + 2u^{\prime}y_1^{\prime} + uy_1^{\prime\prime} \end{align*}\]Substituting these into equation ($\ref{eqn:ode_with_constant_coefficients}$), we get:
\[(u^{\prime\prime}y_1 + 2u^\prime y_1^\prime + uy_1^{\prime\prime}) + a(u^\prime y_1 + uy_1^\prime) + buy_1 = 0\]Rearranging terms with $u^{\prime\prime}$, $u^\prime$, and $u$:
\[y_1u^{\prime\prime} + (2y_1^\prime + ay_1)u^\prime + (y_1^{\prime\prime} + ay_1^\prime + by_1)u = 0\]Since $y_1$ is a solution to equation ($\ref{eqn:ode_with_constant_coefficients}$), the last term in parentheses is 0, and since:
\[2y_1^\prime = -ae^{-ax/2} = -ay_1\]the first term in parentheses is also 0. Therefore, only $u^{\prime\prime}y_1 = 0$ remains, which gives us $u^{\prime\prime}=0$. Integrating twice, we get $u = c_1x + c_2$, and since the 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 for 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 ($\ref{eqn:lambdas}$):
\[\cfrac{1}{2}\sqrt{a^2 - 4b} = \cfrac{1}{2}\sqrt{-(4b - a^2)} = \sqrt{-(b-\frac{1}{4}a^2)} = i\sqrt{b - \frac{1}{4}a^2}\]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 the complex conjugate roots $\lambda = -\cfrac{1}{2}a \pm i\omega$, which correspond to the two complex solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$):
\[\begin{align*} e^{\lambda_1 x} &= e^{-(a/2)x + i\omega x}, \\ e^{\lambda_2 x} &= e^{-(a/2)x - i\omega x} \end{align*}\]However, we can also obtain a basis of real solutions as follows.
Using Euler’s formula:
\[e^{it} = \cos t + i\sin t \label{eqn:euler_formula}\tag{8}\]and substituting $-t$ for $t$ to get:
\[e^{-it} = \cos t - i\sin t\]Adding and subtracting these equations, we obtain:
\[\begin{align*} \cos t &= \frac{1}{2}(e^{it} + e^{-it}), \\ \sin t &= \frac{1}{2i}(e^{it} - e^{-it}). \end{align*} \label{eqn:cos_and_sin}\tag{9}\]The complex exponential function $e^z$ for 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:
\[e^z = e^{r + it} = e^r e^{it} = e^r(\cos t + i\sin t) \label{eqn:complex_exp}\tag{10}\]Setting $r=-\cfrac{1}{2}ax$ 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, adding these equations and multiplying both sides by $\cfrac{1}{2}$, we obtain the first real solution $y_1$:
\[y_1 = e^{-(a/2)x} \cos{\omega x}. \label{eqn:basis_1}\tag{11}\]Similarly, subtracting the second equation from the first and multiplying both sides by $\cfrac{1}{2i}$, we obtain the second real solution $y_2$:
\[y_2 = e^{-(a/2)x} \sin{\omega x}. \label{eqn:basis_2}\tag{12}\]Since $\cfrac{y_1}{y_2} = \cot{\omega x}$ is not constant, $y_1$ and $y_2$ are linearly independent on all intervals and therefore form a basis for the real solutions to equation ($\ref{eqn:ode_with_constant_coefficients}$). From this, we obtain the general solution:
\[y = e^{-ax/2}(A\cos{\omega x} + B\sin{\omega x}) \quad \text{(where }A,\, B\text{ are arbitrary constants)} \label{eqn:general_sol_3}\tag{13}\]