Question 5 of 10
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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