### Differentials and integrals: Applications of integration

### Simple differential equations

Anticipating on instruction about the theory of ordinary differential equations we can already look a special class of differential equations. For now, it suffices to say that a differential equation is an equation in which next to one or more unknown functions also one or more derivatives of those functions are present. The unknown in a differential equation is not a number, but a function (of place, time, or both). A simple example of a differential equation is \[y'(t) = 0\] The unknown function is \(y(t)\) for which the derivative is zero. This is only possible if the function is constant. We call this the general solution of the differential equation.

A direct application of integration is the solution of an ordinary differential equation of type \[\frac{\dd y}{\dd t}=f(t)\] for a given function \(f(t)\). The general solution of this differential equation you get by integration: \[y(t)=\int f(t)\,\dd t\] In other words:

The general solution of \[y'(t)=f(t)\] is \[y(t)=F(t)+c\] where \(F(t)\) is a primitive of \(f(t)\) and \(c\) a constant.

One concrete example is the differential equation \(y'(t)=2t\). Thus, the general solution is \(y(t)=\int 2t\,\dd t=\frac{1}{2}t^2+c\), for some constant \(c\).

You might think that this kind of differential equations is rare, but that is not really the case.

Example from biomechanics In the context of biomechanics one can formulate Newton's second law as the equality of the force \(F\) exerted on an object to the change of the momentum (denoted by \(p\) ) of this object; in formula language \(F=\frac{\dd p}{\dd t}\). If the force \(F(t)\) is known as a function of time \(t\) at an interval \([t_\mathrm{begin}, t_\mathrm{end}]\), then the change in momentum can be calculated during this interval via the following integral: \[p_\mathrm{end}-p_\mathrm{begin}=\int_{t_\mathrm{begin}}^{t_\mathrm{end}} F(t)\,\dd t\]

Example of a jumping upward A concrete example is a jump from the ground upwards with take-off velocity \(v_{\mathrm{take}\textbf{-}\mathrm{off}}\). The time \(t_\mathrm{max}\) needed to reach the maximum height can then be calculated as follows, under the assumption that only gravity \(F(t)=-m\cdot g\), with mass \(m\) and acceleration of gravity \(g\), plays a role and the take-off happens at time \(t=0\): \[-m\cdot v_{\mathrm{take}\textbf{-}\mathrm{off}} =\int_{0}^{t_\mathrm{max}} -m\cdot g\,\dd t=-m\cdot g \cdot t_\mathrm{max}\] In other words, \[v_{\mathrm{take}\textbf{-}\mathrm{off}}=g\cdot t_\mathrm{max}\] the aerial phase lasts twice as long as the time to reach maximum height after take-off: \[\mathrm{duration\;of\;aerial\;phase} = \frac{2v_{\mathrm{take}\textbf{-}\mathrm{off}}}{g}\]

When we discuss the solution method for a so-called separable differential equation, then the theory of integration will fully prove its usefulness.