*There is one differential equation that everybody probably knows, that is Newton’s Second Law of Motion.*If an object of mass \(m\) is moving with acceleration \(a\) and being acted on with force \(F\) then Newton’s Second Law tells us.As you will see most of the solution techniques for second order differential equations can be easily (and naturally) extended to higher order differential equations and we’ll discuss that idea later on.

A linear differential equation is any differential equation that can be written in the following form.

\[\begin \left( t \right)\left( t \right) \left( t \right)\left( t \right) \cdots \left( t \right)y'\left( t \right) \left( t \right)y\left( t \right) = g\left( t \right) \label\end\] The important thing to note about linear differential equations is that there are no products of the function, \(y\left( t \right)\), and its derivatives and neither the function or its derivatives occur to any power other than the first power.

The first definition that we should cover should be that of differential equation.

A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives.

\[\begin F = ma \label \end\] To see that this is in fact a differential equation we need to rewrite it a little.

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First, remember that we can rewrite the acceleration, \(a\), in one of two ways.

\[\beginay'' by' cy = g\left( t \right)\label\end\] \[\begin\sin \left( y \right)\frac = \left( \right)\frac \label\end\] \[\begin 10y''' - 4y' 2y = \cos \left( t \right) \label\end\] \[\begin\frac = \frac\label\end\] \[\begin = \label\end\] \[\begin\frac = 1 \frac \label\end\] The order of a differential equation is the largest derivative present in the differential equation.

In the differential equations listed above \(\eqref\) is a first order differential equation, \(\eqref\), \(\eqref\), \(\eqref\), \(\eqref\), and \(\eqref\) are second order differential equations, \(\eqref\) is a third order differential equation and \(\eqref\) is a fourth order differential equation.

\[\begina = \frac\hspace\hspace\,\,\,\,\,\,a = \frac \label\end\] Where \(v\) is the velocity of the object and \(u\) is the position function of the object at any time \(t\).

We should also remember at this point that the force, \(F\) may also be a function of time, velocity, and/or position.

## Comments How To Solve An Initial Value Problem

## Differential Equations - WolframAlpha Examples

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## Solving an initial value problem with ode45 - UMD MATH

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