First check the characteristic solution:
<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0
has characteristic equation
<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0
with a double root at <em>r</em> = -2, so the characteristic solution is

For the particular solution corresponding to
, we might first try the <em>ansatz</em>

but
and
are already accounted for in the characteristic solution. So we instead use

which has derivatives


Substituting these into the left side of the ODE gives

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.
For the second solution corresponding to
, we use

with derivative


Substituting these gives

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.
Then the general solution to the ODE is

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have


and so the particular solution satisfying these conditions is

or
