Here we focus on solving first order homogeneous linear differential equation: \[ \dot{x}(t) +p(t)x(t) = 0 \tag{1} \] where overdot indicates derivative with respect to \(t\). Assume the boundary (or initial) condition of the equation is \(x(0)=x_0\). Solving this equation is straightforward since it is separable, i.e., the variables can be separated to find the solution of the equation: \[ \begin{align*} \frac{dx}{dt} &= -p(t)x(t) \\ \int_{x(0)}^{x(t)}\frac{dx}{x} &= -\int_{0}^{t} p(\tau)d\tau \\ \ln x(t) - \ln x(0) &= -\int_{0}^{t} p(\tau)d\tau \;\; \Longrightarrow \;\; x(t) = x(0)\exp\bigg( -\int_{0}^{t}p(\tau)d\tau \bigg) \end{align*} \] In case if \(p(t)\) is a constant function, \(p(t)=-a\), then we have: \[ x(t) = x(0)\exp(at) \tag{2} \]
Here we focus on solving first order non-homogeneous linear differential equation: \[ \dot{x}(t) +p(t)x(t) = f(t) \tag{3} \] where we now have a non-zero function \(f(t)\) on the right-hand side. Again, we assume the boundary (or initial) condition of the equation is \(x(0)=x_0\).
A common way to solve this equation is to multiply an integrating factor on both sides. In detail, let us denote the integrating factor as \(I(t)\). If we choose \(I(t)\) to satisfy \(\dot{I}(t)=I(t)p(t)\), then:
\[ \begin{align*} I(t)\dot{x}(t) +\{I(t)p(t)\}x(t) &= I(t)f(t) \\ \frac{d}{dt} \bigg( I(t)x(t) \bigg) &= I(t)f(t) \\ I(t)x(t) - I(0)x(0) &= \int_{0}^{t} I(\tau)f(\tau)d\tau \tag{4} \end{align*} \] It is clear that the integrating factor itself constitutes a differential equation: \[ \dot{I}(t)=I(t)p(t) \; \Longrightarrow \; I(t) = I(0)\exp\bigg(\int_{0}^{t}p(\tau)d\tau\bigg) \] Putting this result back to equation (4), we have: \[ \begin{align*} x(t) &= x(0)\exp\bigg(-\int_{0}^{t}p(\tau)d\tau\bigg) + \int_{0}^{t}\exp\bigg(\int_{0}^{\tau}p(\tau')d\tau' -\int_{0}^{t}p(\tau')d\tau' \bigg)f(\tau)d\tau \\ &= x(0)\exp\bigg(-\int_{0}^{t}p(\tau)d\tau\bigg) + \int_{0}^{t}\exp\bigg(-\bigg[ \int_{\tau}^{0}p(\tau')d\tau' +\int_{0}^{t}p(\tau')d\tau' \bigg] \bigg)f(\tau)d\tau \\ &= x(0)\exp\bigg(-\int_{0}^{t}p(\tau)d\tau\bigg) + \int_{0}^{t}\exp\bigg( -\int_{\tau}^{t}p(\tau') f(\tau)d\tau \bigg) \end{align*} \] If we again, consider the case when \(p(t)=-a\), then: \[ \begin{align*} x(t) = x(0)\exp(at) + \int_{0}^{t}\exp\big(a(t-\tau)\big)f(\tau)d\tau \tag{5} \end{align*} \] Note that by definition, the final term of equation (5) is a convolution between function \(\exp(at)\triangleq g(t)\) and \(f(t)\): \[ [g * f](t) \triangleq \int_{0}^{t} g(t-\tau)f(\tau)d\tau = \int_{0}^{t}\exp\big(a(t-\tau)\big)f(\tau)d\tau \]
If we only consider cases when \(p(t)\) is a constant function, i.e., \(p(t)=-a\). \[ \dot{x}(t) = ax(t) \tag{6} \] Then, there are other approaches to solve the first order homogeneous linear differential equation.
Assume that the solution of \(x(t)\)
can be expressed as an infinite sum of polynomials: \[
x(t) = c_0 + c_1 t + c_2 t^2 + \cdots = \sum_{k=0}^{\infty} c_k t^{k}
\tag{7}
\]
putting equation (7) into (6) provides us: \[
c_1 + 2 c_2t + 3c_3 t^2 \cdots = ac_0 + ac_1 t + ac_2 t^2 + \cdots
\] and hence we have the following recurrance relation: \[
\forall k \in \mathbb{N}^{+}, \;\; (k+1)c_{k+1} = ac_{k}
\] where \(\mathbb{N}^{+}\)
stands for a set consisting of non-negative integers. Using the boundary
condition \(x(0)=c_0\), we have: \[
c_{k} = \frac{a^k}{k!}x(0)
\] and putting this solution back to equation (7) results in
equation (2). \[
x(t) = x(0) \sum_{k=0}^{\infty} \frac{(at)^{k}}{k!} = x(0)\exp(at)
\]
Consider a vector space \((F,+,\cdot)\), where \(F\) is set of functions that maps from a real number is and the operations are defined are defined pointwise [REF]. The differential operator can be regarded as a linear operator between function vector spaces. Hence, we are finding the eigenfunction under the differential operator with eigenvalue \(a\).