30.4. HIGHER ORDER SCALAR LINEAR EQUATIONS 597

Then using the above theorem,xp (t) =

∫ t

0

 −eu (2eu−3) 0 −6eu (eu−1)−eu (u+ eu−1) eu −eu (2u+3eu−3)

eu (eu−1) 0 eu (3eu−2)



t−ut−u

ln((t−u)2 +1

)du

This gives a perfectly good description of a particular solution. Thus the general solutionis of the form  −et (2et −3) 0 −6et (et −1)

−et (t + et −1) et −et (2t +3et −3)et (et −1) 0 et (3et −2)

c+xp (t)

Here c is an arbitrary vector in Fn. Note how this is essentially a return to the notion of thevariation of constants formula presented earlier.

Of course all of this depends on being able to say that if two functions have the sameLaplace transform, then they must in some sense be the same function. This will be dis-cussed later when it will also be shown how to explicitly go backwards in the table and findthe original function given its Laplace transform.

30.4 Higher Order Scalar Linear EquationsRecall these are differential equations which are of the form

y(n)+an−1 (t)y(n−1)+ · · ·+a1 (t)y′+a0 (t)y = f (t)

The notation y(k) means the kth derivative, and it is assumed that all given functions arecontinuous. There is nothing new about these equations. They can all be studied as specialcases of first order systems.

The following is a procedure for changing one of these higher order linear equationsinto a first order system which is the right way to study differential equations.

PROCEDURE 30.4.1 Consider the equation

y(n)+an−1 (t)y(n−1)+ · · ·+a1 (t)y′+a0 (t)y = f

To write as a first order system, do the following.x(1)x(2)

...x(n)

=

yy′

...y(n−1)

Then, suppressing the dependence on t,

x(1)x(2)

...x(n−1)

x(n)



′

=



x(2)x(3)

...x(n)y(n)

=



x(2)x(3)

...x(n)

−(an−1x(n)+ · · ·+a1x(2)+a0x(1))+ f

