The auxiliary equation for our differential equation is:

A.E. `m^2-4m+4` `=(m-2)^2` `=0`

In this case, we have:

`m=2` (repeated root)

We need to use the second form from the table above (`y = e^(mx)(A + Bx)`), and once again use the correct variables (*t* and *s*, instead of *x* and *y*).

So `s(t)=(A+Bt)e^(2t)`

Now to find the values of the constants:

`s(0)=1` implies `A=1`

So we can write

`s(t)=(1+Bt)e^(2t)`

Differentiating gives:

`s'(t)=(1+Bt)2e^(2t)+Be^(2t)`

Now

`s'(0)=3 ` implies `2+B=3`

This gives us `B=1`

So

`s(t)=(1+t)e^(2t)`

The graph of our solution is as follows:

Graph of displacement `s(t)=(1+t)e^(2t)`.