Mathematical Tools for Physics

4—Differential Equations 89

The homogeneous part of Eq. ( 11 ) has the solution found in Eq. ( 6 ) and the total is

x(t) =xhom(t) +xinh(t) =x(t) =Aeα^1 t+Beα^2 t+F 0

(

1

k

−

1

mβ^2 −bβ+k

e−βt

)

There are two arbitrary constants here, and this is what you need because you have to be able to specify the
initial position and the initial velocity independently; this is a second order differential equation after all. Take for
example the conditions that the initial position is zero and the initial velocity is zero. Everything is at rest until
you start applying the external force. This provides two equations for the two unknowns.

x(0) = 0 =A+B+F 0

mβ^2 −bβ

k(mβ^2 −bβ+k)

x ̇(0) = 0 =Aα 1 +Bα 2 +F 0

β

mβ^2 −bβ+k

Now all you have to do is solve the two equations in the two unknownsAandB. I would take the first, multiply
it byα 2 and subtract the second. This givesA. Do the same withα 1 instead ofα 2 to getB. The results are

A=

1

α 1 −α 2

F 0

α 2 (mβ^2 −bβ)−kβ

k(mβ^2 −bβ+k)

Interchangeα 1 andα 2 to getB.
The final result is

x(t) =

F 0

α 1 −α 2

(

α 2 (mβ^2 −bβ)−kβ

)

eα^1 t−

(

α 1 (mβ^2 −bβ)−kβ

)

eα^2 t

k(mβ^2 −bβ+k)

+F 0

(

1

k

−

1

mβ^2 −bβ+k

e−βt

)

(13)

If you think this is messy and complicated, you haven’t seen messy and complicated. When it takes 20
pages to write out the equation, then you’re entitled say that it’s starting to become involved.
The problem isn’t finished until you’ve analyzed the supposed solution. After all, I may have made some
errors in algebra along the way. Also, analyzing the solution is the way you learn how these functionswork.