The study of differential equations often begins with the analysis of the simplest yet most ubiquitous case: the first-order linear homogeneous ordinary differential equation with constant coefficients.
This equation models systems where the rate of change of a quantity is directly proportional to the quantity itself.
Its mathematical formulation is given by an initial value problem (IVP), considering a function x(t) with its initial state at t=0 specified as x(0) = x_0.
The governing equation is:
\frac{\mathrm{d}x}{\mathrm{d}t} = \lambda x(t)
Here, \lambda is a real or complex constant that determines the nature of the change, whether it be exponential growth (\lambda > 0), decay (\lambda < 0), or oscillation (if \lambda is imaginary).
We will explore two analytical methods to obtain the solution to this IVP, demonstrating a connection between differential equations and the definition of the exponential function.
The method of separation of variables is a direct approach applicable when the equation can be algebraically rearranged so that all terms involving the dependent variable x are on one side and all terms involving the independent variable t are on the other.
First, we address the trivial case. If the initial condition is x_0 = 0, one may verify by inspection that the function x(t) = 0 for all t is a solution. The derivative is \frac{\mathrm{d}x}{\mathrm{d}t} = 0, and the right-hand side is \lambda \cdot 0 = 0, satisfying the equation.
For the non-trivial case where x_0 \neq 0, the solution x(t) will be non-zero in some neighborhood of t=0 by continuity.
In this neighborhood, we can divide by x(t):
\frac{1}{x(t)} \frac{\mathrm{d}x}{\mathrm{d}t} = \lambda
We can then integrate both sides with respect to t from the initial time t=0 to an arbitrary time t.
To formalize this, we employ a change of variables on the left-hand side:
\int_0^t \frac{1}{x(\tau)} \frac{\mathrm{d}x}{\mathrm{d}\tau} \mathrm{d}\tau = \int_0^t \lambda \, \mathrm{d}\tau
Performing the integration yields:
\begin{aligned} & \int_{x(0)}^{x(t)} \frac{1}{\xi} \, \mathrm{d}\xi = \int_0^t \lambda \, \mathrm{d}\tau \\ & \ln|\xi| \bigg|_{x_0}^{x(t)} = \lambda\tau \bigg|_0^t \\ & \ln|x(t)| - \ln|x_0| = \lambda t \\ & \ln\left| \frac{x(t)}{x_0} \right| = \lambda t \end{aligned}
Taking the exponential of both sides:
\left| \frac{x(t)}{x_0} \right| = e^{\lambda t}
Since e^{\lambda t} is always positive, the ratio \frac{x(t)}{x_0} must be either consistently positive or consistently negative. As the solution must be continuous and starts at t=0 with a ratio of 1, it must remain positive for all t:
\frac{x(t)}{x_0} = e^{\lambda t}
This leads to the unique solution for the initial value problem:
x(t) = x_0 \, e^{\lambda t}
This solution also incorporates the trivial case where x_0=0.
An alternative and more abstract approach is to seek a solution in the form of a power series expansion around the initial point t=0.
This technique does not presuppose the form of the solution and can be used to define the exponential function itself. We search for a solution of the form:
x(t) = \sum_{n=0}^{\infty} C_n t^n
where the coefficients C_n are to be determine and we assume the series converges on some interval |t| < R.
Within this interval, we can differentiate the series term-by-term:
\frac{\mathrm{d}x}{\mathrm{d}t} = \sum_{n=1}^{\infty} n C_n t^{n-1}
Substituting these series representations into the original differential equation \dot{x} = \lambda x:
\sum_{n=1}^{\infty} n C_n t^{n-1} = \lambda \sum_{n=0}^{\infty} C_n t^n
To compare the coefficients, we must align the powers of t. We perform an index shift on the left-hand side summation by letting k = n-1. Consequently, n=k+1, and the summation starts from k=0:
\begin{aligned} & \sum_{k=0}^{\infty} (k+1) C_{k+1} t^k = \lambda \sum_{k=0}^{\infty} C_k t^k \\ & \sum_{k=0}^{\infty} \left[ (k+1) C_{k+1} - \lambda C_k \right] t^k = 0 \end{aligned}
According to the identity theorem for power series, if this equation holds for all t in an open interval, then the coefficient of each power of t must be zero.
This yields a recurrence relation for the coefficients:
\begin{aligned} & (k+1) C_{k+1} - \lambda C_k = 0 \quad \implies \\ & C_{k+1} = \frac{\lambda}{k+1} C_k \quad \text{for } k \ge 0 \end{aligned}
The first coefficient, C_0, is determined from the initial condition:
x(0) = \sum C_n (0)^n = C_0
Therefore, C_0 = x_0. We can now generate the subsequent coefficients recursively:
\begin{aligned} C_1 &= \frac{\lambda}{1} C_0 = \lambda x_0 \\ C_2 &= \frac{\lambda}{2} C_1 = \frac{\lambda^2}{2 \cdot 1} C_0 = \frac{\lambda^2}{2!} x_0 \\ C_3 &= \frac{\lambda}{3} C_2 = \frac{\lambda^3}{3 \cdot 2 \cdot 1} C_0 = \frac{\lambda^3}{3!} x_0 \\ &\vdots \\ C_k &= \frac{\lambda^k}{k!} C_0 = \frac{\lambda^k}{k!} x_0 \end{aligned}
Substituting this general form for the coefficients back into the power series for x(t):
x(t) = \sum_{n=0}^{\infty} \left(\frac{\lambda^n x_0}{n!}\right) t^n = x_0 \sum_{n=0}^{\infty} \frac{(\lambda t)^n}{n!}
The resulting series is the Maclaurin series for the exponential function, which is by definition:
e^z = \sum_{n=0}^\infty \frac{z^n}{n!}
Identifying z = \lambda t, we arrive at the same solution:
x(t) = x_0 \, e^{\lambda t}
The validity of the term-by-term differentiation is justified by the convergence of the series.
Applying the ratio test, we find that the radius of convergence is infinite, confirming that this solution is valid for all t \in \mathbb{R}.
This method not only solves the equation but also demonstrates that the exponential function is the natural solution to the simplest law of change, providing a deep link between calculus and the fundamental constants of mathematics.
The uniqueness of this solution is guaranteed by the Picard–Lindelöf theorem (here), as the function f(t,x) = \lambda x is Lipschitz continuous with respect to x over the entire domain.