Higher order Linear ODEs
Contents
Higher order Linear ODEs#
Wronskian#
We define a Wronskian of two functions as the determinant
Now consider the Wronskian of two solutions of a homogeneous ODE \(y'' + p(x)y' + q(x)y = 0\):
Differentiating it
In other words, the Wronskian satisfies the first-order linear ODE
whose solution is the Abel’s identity
In general, the Wronskian of a set of functions \(\{f_1(x), f_2(x), \dots, f_n(x) \}\) is defined by
Second-order linear ODE#
Reduction of Order - homogeneous#
Often we can find one solution of an ODE from inspection, which we can then use to find the other linearly independent solution.
And integrating
Can set C = 0 since we write the solution as a linear combination of u1 u2 anyway
Variation of parameters - inhomogeneous#
Let the general solution of the corresponding homogeneous equation be known
As we have done before, we use variation of parameters to set the ansatz for the particular solution:
where we have “promoted” the constants \(C_1, C_2\) to functions \(A(x), B(x)\). Differentiating it once:
We also require that
so eq. (156) becomes
Differentiating again
Now we substitute (159) and (160) into eq. (155):
We get
The solution of the system of equations (157) and (161) is given by
and after integrating we get
General \(n\)-th order linear ODE#
We can directly extend the method of variation of parameters we used to solve second-order linear ODE to \(n\)-th order ODEs, which are of the form
Let \(\{y_1, y_2, \dots, y_n\}\) be solutions of the homogeneous ODE, i.e. \(\mathcal{L}[y_i] = 0\). We seek solutions of the inhomogeneous ODE in the form of ansatz
where \(C_i(x)\) are functions that we need to find. As before, we differentiate and define conditions:
These \(n\) equations define a system of linear equations with unknowns \(C_1, \dots, C_n\):
A sufficient condition for the existence of a solution of the system of equations (128) is that the determinant of coefficients is nonzero for each value of \(x\). This is equal to the Wronskian \(W[y_1, \dots, y_n]\) and we know it is nonzero for all \(x\) because the set \(\{y_1, \dots, y_n \}\) is linearly independent. We determine \(C'_1, \dots, C'_n\) using Cramer’s rule:
where \(W_i\) is the determinant obtained from \(W\) by replacing \(i\)-th column with the RHS column \((0, 0, \dots, r)^T\). We integrate this and substitute the result in eq. (127) to get the particular solution of (186)
where \(x_0\) is arbitrary.
Linear ODEs with Constant Coefficients#
In this section we focus on second-order ODEs where functions \(p(x), q(x)\) and \(r(x)\) are now constants. We therefore write Eq. inhmg2ndode
as
Homogeneous case - method of characteristic polynomial#
We use the method of characteristic polynomial which can be used for any such ODE of any order. A homogeneous linear ODE of first order with constant coefficient \(a_0\)
is separable and its integral is \(y(x) = Ce^{\lambda x}\), where \(\lambda\) is an unknown constant. We find it by plugging \(y\) back into the ODE:
This is the characteristic equation of the ODE. We find \(\lambda = -a_0\).
Consider now a second-order ODE with constant coefficients \(a_1, a_0\)
Inspired by the solution of the first-order equation, we assume the solution of the form \(y(x) = Ce^{\lambda x}\). We plug it back into the ODE and get the characteristic equation
The roots of this quadratic equation are
so we have three cases depending whether \(D > 0, D = 0\) or \(D < 0\), keeping in mind that we need linearly independent solutions to form the basis:
\( \lambda_{1,2} \in \mathbb{R}, \lambda_1 \neq \lambda_2: \quad y(x) = C_1 e^{\lambda_1 x} + C_2e^{\lambda_2 x} \)
\( \lambda_{1,2} \in \mathbb{R}, \lambda_1 = \lambda_2: \quad y(x) = (C_1 + C_2 x)e^{\lambda_1 x} \)
\( \lambda_{1,2} \in \mathbb{C}, \lambda_2 = \lambda_1^: \quad y(x) = C_1 e^{\lambda_1 x} + C_2e^{\lambda_1^ x} \)
Hint: Diagonalisation
The method of characteristic polynomial might remind us of the process of finding eigenvalues. Indeed, we can think of finding solutions in the form of an exponential function as a type of diagonalisation.
The exponential function can be thought of as an eigenvector of the derivative operator and \(\lambda\) as the eigenvalue. If \(\lambda_1, \lambda_2\) are roots of the characteristic equation, we can write
Let us now generalise this to an \(n\)-th order ODE. The characteristic equation of eq. (129) is again obtained by substituting \(y = e^{\lambda x}\):
For any distinct real root \(\lambda\), one solution is \(y = e^{\lambda x} \)
For any complex root \(\lambda = \gamma + i \omega\), its conjugate \(\lambda^* = \gamma - i \omega\) is also a root:
Multiple real roots: if a real root is repeated \(m\) times, the \(m\) corresponding linearly independent solutions are
Multiple complex roots: if a complex root \(\lambda = \gamma + i \omega\) is repeated, the corresponding linearly independent solutions are