# General solution of second-order linear ODE

I am trying to look a bit deeper into the mathematics the equation of motion used in physics and engineering. I have some specific questions at the end, but please correct me if I make a mistake in my statement.

I have an ordinary, second-order, linear, homogeneous differential equation:

$$M \ddot{y} + C \dot{y} + K y = 0$$

where $M$, $C$ and $K$ are real valued.

I am familiar with an Ad-hoc method of solving this equation. Namely, I assume a solution takes form of

$$y_k(t) = Y_k e^{\omega_k t}$$

substitute the form into the ODE,

$$\left( M \omega_k^2 + C \omega_k + K\right) y_k = 0,$$

and solve for $\omega_k$ such that $y_k \neq 0$.

If $y$ is in dimension $N$, we generally have $2N$ solutions, where for each $k$, $\omega_k$ and its complex conjugate $\tilde{\omega}_k$ are solutions. Since our ODE is linear, the general solution is a combination of the individual solutions

$$y(t) = \sum_{k = 1}^N Y_k e^{\omega_k t} + \tilde{Y}_k e^{\tilde{\omega}_k t}$$

The values of $Y_k$ and $\tilde{Y}_k$ are determined from initial values, i.e. $y(0) = y_0$, and further restrictions common in physics, such as $y(t)$ must be real for all $t \geq 0$.

Now here are my questions:

1 - If we select the form $y_k(t) = Y_k e^{ \mathbf{i}\omega_k t}$, am I correct to say that the roots $\omega_k$ and $\tilde{\omega}_k$ are no longer complex conjugates, but $-1$ times their complex conjugates?

2 - How do we know that the general solution cannot contain terms which are not representable by the exponential form? In other words, is our general solution truly general?

• Just a remark: when you write "$M$, $C$ and $K$ are real valued", you probably mean "$M$, $C$ and $K$ are real constants". Am I right? – Siminore Aug 11 '12 at 13:20
• Yes, you are correct. – Indrid Cold Aug 11 '12 at 13:27

## 3 Answers

Here is another approach to the second part of your question:

Let $x_1 = \dot{y}, x_2 = y$. Assuming $M \neq 0$, you can write your equation as a pair of first order equations: $$\dot{x_1} = -M^{-1} (C x_1 + K x_2), \ \ \ \dot{x_2} = x_1, \ \ \ x_1(0) = \dot{y}(0), \ \ \ x_2(0) = y(0).$$ This is a time-invariant linear system, so if we let $x = (x_1,x_2)^T \in \mathbb{R}^{2 }$, then the equation can be written as: $$\dot{x} = A x = M^{-1}\begin{bmatrix} -C & -K \\ 1 & 0 \end{bmatrix} x, \ \ \ x(0) = x^0 = (\dot{y}(0), y(0))^T.$$ By the existence and uniqueness theorem for ODEs (this is a classical result, for example see Marsden, "Elementary Classical Analysis", Theorem 7.6), the system has a unique solution passing through the point $x^0 \in \mathbb{R}^{2}$.

Hence the initial condition $x^0$ completely defines the solution. Since this system is linear, the solutions are linear functions of the initial state, which simplifies matters considerably. We need only consider solutions starting from a basis of $\mathbb{R}^2$. In particular, all solutions are linear combinations of these two solutions (because of linearity).

To see why these two solutions have exponentials, you need to consider two cases:

(1) If $A$ is diagonalizable, then the eigenvectors of $A$ can form a basis of $\mathbb{R}^2$, and it is straightforward to check that $t \mapsto e^{\lambda_i t} v_i$ are solutions, where $v_i$ are the eigenvectors corresponding to the eigenvalues $\lambda_i$.

(2) If $A$ is not diagonalizable, then it has only one eigenvalue, say $\lambda$, and we can find a basis $u,v$ of $\mathbb{R}^2$ that satisfies $A v = \lambda v$ and $A u = \lambda u +v$ (cf. the Jordan form). Then it is straightforward to check that $t \mapsto e^{\lambda t}v$ and $t \mapsto e^{\lambda t} (u + tv)$ are solutions.

Hence all solutions are of exponential form (including, of course, multiplying by $t$ as well).

About your second question, the solutions of a homogeneous linear ODE is a vector space of precise dimension, so you should discover a precise number of linearly independent solutions. No matter how you construct them, if you find the correct number of linearly independent solutions, the general solution is a linear combinations of these generators. This is the most important result about linear ODEs with constant coefficients.

Edit. Just a sketch of the proof for a second order equation. Consider the differential operator $$Lu=au''+bu'+cu,$$ $a$, $b$ and $c$ being fixed constants. Pick a point $t_0 \in \mathbb{R}$ and consider the two Cauchy problems $$\left\{ \begin{array}{ll} Lu=0 \\ u(t_0)=0 \\ u'(t_0)=1 \end{array} \right. \qquad \left\{ \begin{array}{ll} Lu=0 \\ u(t_0)=1 \\ u'(t_0)=0 \end{array} \right.$$ Both problems have a unique, globally defined solution, by a fundamental resultin the theory of ODEs. Call them $u_1$ and $u_2$. It is now a simple exercise to check that any function $w$ such that $Lw=0$ can be written in the form $w=c_1u_1+ c_2 u_2$ for suitable values of $c_1$ and $c_2$. Hence $\{u_1,u_2\}$ generates the set of solutions to $Lu=0$. Since $u_1$ and $u_2$ are linearly independent (another simple exercise), $\{u_1,u_2\}$ is a basis of the kernel of $L$.

• Siminore, intuitively I agree with your statement, but how do I show that the solution of the ODE is a vector-space of 2N dimensions? I understand this is probably a fundamental theorem, can you suggest a reference? – Indrid Cold Aug 11 '12 at 13:35
• @IndridCold The most important tool is the existence and uniqueness theorem for the Cauchy problem. I'll add an edit to my answer. – Siminore Aug 11 '12 at 14:10

The second question has been already answered by @Siminore.

Regarding the first question, if you have solutions $\alpha\pm i\beta$ when using $e^{\omega t}$, then when using $e^{i\lambda t}$ you have $$\lambda=-i\omega=-i(\alpha\pm i\beta)=-i\alpha\pm\beta$$ so it is the real part that changes sign.

• Thank you enzotib. I have seen this detail missed in a number of engineering references. Physically, this detail makes a huge difference. – Indrid Cold Aug 11 '12 at 13:38