Editing Integral equation (section)

== Converting IVP to integral equations ==
In the following section, we give an example of how to convert an initial value problem (IVP) into an integral equation. There are multiple motivations for doing so, among them being that integral equations can often be more readily solvable and are more suitable for proving existence and uniqueness theorems.<ref name=":1" />

The following example was provided by Wazwaz on pages 1 and 2 in his book.<ref name=":0" /> We examine the IVP given by the equation:

<math display="block">u'(t) = 2tu(t), \, \, \,\,\, \,\, x \geq 0 </math>and the initial condition:

<math display="block">u(0)=1</math>

If we integrate both sides of the equation, we get:

<math display="block">\int_{0}^{x}u'(t) \, dt = \int_0^x 2tu(t) \, dt</math>

and by the [[fundamental theorem of calculus]], we obtain:

<math display="block">u(x)-u(0) = \int_0^x 2tu(t) \, dt</math>

Rearranging the equation above, we get the integral equation:

<math display="block">u(x)= 1+ \int_0^x 2tu(t) \, dt</math>

which is a Volterra integral equation of the form:

<math display="block">u(x) = f(x) + \int_{\alpha(x)}^{\beta(x)}K(x,t) \cdot u(t) \, dt</math>

where ''K''(''x'',''t'') is called the kernel and equal to 2''t'', and ''f''(''x'')&nbsp;=&nbsp;1.<ref name=":0" />