Introduction

Ordinary differential equations: Second-order linear ODEs with constant coefficients

Introduction

Second-order linear differential equations, which are of interest in science and technology, can be written in the form $a(t)\cdot\frac{\dd^2y}{\dd t^2}+b(t)\cdot \frac{\dd y}{\dd t}+c(t)\cdot y=f(t)$ If $f(t)=0$ then the ODE is called homogeneous. The functions $a(t)$ , $b(t)$ and $c(t)$ are called coefficients. When these coefficients are constant, then the general solution can be expressed in terms of elementary functions. This is the only type of second-order linear differential equations that we will deal with, but even then there already many applications. For inspiration, we start with the most famous example: the so-called mathematical pendulum.

The general solution of the mathematical pendulum, that is, of the ODE $\frac{\dd^2y}{\dd t^2}+\omega^2y=0$ where $\omega$ is some real parameter is $y(t)=A\cos(\omega t)+B\sin(\omega t)$ where $A$ and $B$ are constants, which are in most cases fixed when two additional conditions are specified. These may be two function values or one function value and one derivative at a point.

We consider the following second-order differential equation of the mathematical pendulum $\frac{\dd^2y}{\dd t^2}+\omega^2y=0$ where $\omega$ is some fixed number. You can try to guess a solution. A suitable guess could be a solution of the form $y(t)=e^{\lambda t}$ where $\lambda$ is a yet to be determined constant. Subsitution in the ODE gives the following condition for $\lambda$ : $\lambda^2 + \omega^2=0$ There are two complex solutions of this equation for $\lambda$ : $\lambda=\mathrm{i}\,\omega$ and $\lambda=-\mathrm{i}\,\omega$ . So, $y_1(t)=e^{\mathrm{i}\,\omega\,t}$ and $y_2(t)=e^{-\mathrm{i}\,\omega\,t}$ are complex solutions of the given ODE. But any linear combination $c_1\cdot y_1(t)+c_2\cdot y_2(t)$ (with complex constants $c_1$ and $c_2$ ) is also a solution. This leads to two real solutions of the ODE, namely $y_c(t)=\frac{e^{\,\mathrm{i}\,\omega\,t}+e^{-\mathrm{i}\,\omega\,t}}{2}=\cos(\omega t)$ and $y_s(t)=\frac{e^{\,\mathrm{i}\,\omega\,t}-e^{-\mathrm{i}\,\omega\,t}}{2\mathrm{i}}=\sin(\omega t)$ Also, each linear combination of these two functions, a solution of the ODE. Of course we could have guessed immediately that the sine and cosine function with angular frequency $\omega$ are solution, taking the standard derivatives of trigonometric functions into account.

We omit the proof that we have found in the above way all solutions of the mathematical pendulum, but it is similar to the proof of the second theorem on this theory page.

The general solution of the ODE $\frac{\dd^2y}{\dd t^2}-\omega^2y=0$ is $y(t)=A\cdot e^{\omega\,t}+B\cdot e^{-\omega\,t}$ where $A$ and $B$ are constants, which are in most cases fixed when two additional conditions are specified. These may be two function values or one function value and one derivative at a point.

Consider the following second-order differential equation $\frac{\dd^2y}{\dd t^2}-\omega^2y=0$ where $\omega$ is some fixed number. Similarly to the solving of the ODE of the mathematical pendulum we guessing again a solution of the form $y(t)=e^{\lambda t}$ Substitution in the ODE gives the following condition for $\lambda$ : $\lambda^2 - \omega^2=0$ There are two solutions of this equation $\lambda$ : $\lambda=\omega$ and $\lambda=-\omega$ . So, $y_1(t)=e^{\omega\,t}$ and $y_2(t)=e^{-\omega\,t}$ are solutions of the given ODE. But any linear combination $c_1\cdot y_1(t)+c_2\cdot y_2(t)$ (with constants $c_1$ and $c_2$ ) is a solution.

Thus, the general solution of the ODE $\frac{\dd^2y}{\dd t^2}-\omega^2y=0$ is $y(t)=A\cdot e^{\omega\,t}+B\cdot e^{-\omega\,t}$ where $A$ and $B$ are constants, which are in most cases fixed when two additional conditions are specified. These may be two function values or one function value and one derivative at a point.

The proof that we have found in the above way all solutions of the given ODE boils down to the proof that for a solution of the form $y(t)=z(t)\cdot e^{\omega t}$ it must be true that $z'(t)+ 2\,\omega\, z(t) = c$ for some constant $c$ . This is just application of rules for differentiation and formula manipulation. We already know the general solution of the latter ODE: $z(t)=A+B\cdot e^{-2\omega t}$ with $A=\dfrac{c}{2\omega}$ and $B$ some constant. Then: $y(t)=A\cdot e^{\omega\,t}+B\cdot e^{-\omega\,t}$ where $A$ and $B$ are constants.