Integro-differential equation

Consider the following second-order problem,

${\displaystyle u'(x)+2u(x)+5\int _{0}^{x}u(t)\,dt=\theta (x)\qquad {\text{with}}\qquad u(0)=0,}$

where

${\displaystyle \theta (x)=\left\{{\begin{array}{ll}1,\qquad x\geq 0\\0,\qquad x<0\end{array}}\right.}$

is the Heaviside step function. The Laplace transform is defined by,

${\displaystyle U(s)={\mathcal {L}}\left\{u(x)\right\}=\int _{0}^{\infty }e^{-sx}u(x)\,dx.}$

Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro-differential equation is converted into the following algebraic equation,

${\displaystyle sU(s)-u(0)+2U(s)+{\frac {5}{s}}U(s)={\frac {1}{s}}.}$

Thus,

${\displaystyle U(s)={\frac {1}{s^{2}+2s+5}}}$.

Inverting the Laplace transform using contour integral methods then gives

${\displaystyle u(x)={\frac {1}{2}}e^{-x}\sin(2x)\theta (x)}$.

Alternatively, one can complete the square and use a table of Laplace transforms ("exponentially decaying sine wave") or recall from memory to proceed:

${\displaystyle U(s)={\frac {1}{s^{2}+2s+5}}={\frac {1}{2}}{\frac {2}{(s+1)^{2}+4}}\Rightarrow u(x)={\mathcal {L}}^{-1}\left\{U(s)\right\}={\frac {1}{2}}e^{-x}\sin(2x)\theta (x)}$.

Integro-differential equations have found applications in epidemiology, the mathematical modeling of epidemics, particularly when the models contain age-structure[2] or describe spatial epidemics.[3]