Linear Differential Equations
Let's start at the begining and say that a differential equation is and equation that contains derivatives. So consider one of the most famous equations, Newton's 2nd law: \[F=ma\] Now \(a\) is an acceleration and we can write \(a={dv \over dt}\) or \(a={d^2s \over dt^2}\) where \(v\) is a velocity and \(s\) is a position function. So, assuming that \(F\) might vary with time, position and/ot velocity we can modify and rearrange the law to get: $$ m{d^2s \over dt^2} = F \left ( t,s,{ds \over dt} \right )$$
The order of the differential equation is just the 'size' of the largest derivative present - in this case we have a second order differential equation. (Remeber that we can use the 'prime' notation - \(dx \over dt\) can be written as \(x'\) and \({d^2x \over dt^2}\) can be written as \(x''\) A Linear Differential Equation (LDE) is a differential equation where the variable in the equation and any of its derivatives are combined linearly. Probably easiest to look at a few examples"
\(x''' + 2x'-23x +.005 = 0\) is linear but \(x'' - cos(x) = 0\) isn't.
\(x'' + x + 3 = sin(t)\) is linear and so is \(y'''-2y't^2 = t\) but \(x' -ln(x)=0\) isn't.
So the general form of an nth order LDE is:
\[a_n(t) {d^nx \over dt^n} + a_{n-1}(t) {d^{n-1}x \over dt^{n-1}} +…+a_1(t) {dx \over dt}+a_0x = g(t)\]
Looking at the solutions for a second order LDE such as:
\[{{{d^2 x} \over {d t^2}} + 3 {{dx} \over {dt}} +2x}= 0\]
If we assume an exponential solution exists such as:
\[x(t) = e^{st}\]
substituting for
\[{{dx} \over {d t} }= {se^{s t}}\]
and for
\[{{d^2x} \over {dt^2} }={ s^2e^{st}}\] we get: \[{e^{st}(s^2 + 3s + 2) }= 0\]
Now, as \(e^{st}\) can never be 0, the solution can exist if, and only if, the polynomial part is 0 The roots of the polynomial bit of the equation are
\[s=-2, -1\]
and the solution of the LDE being a superposition of the solutions for s gives:
\[x(t) = ae^{-2t} + be^{-t}\]
