The idea is as follows. If we have [imath] (A-\lambda I)(x)=0, [/imath] i.e. [imath] Ax=\lambda x [/imath] then we solve [imath] y'=\lambda y [/imath] and obtain [imath] y=me^{\lambda t}. [/imath] However, we actually have for [imath] \lambda=-3 [/imath] that
[math] (A-\lambda I)^2(x)=0. [/math] That means that either [imath] (A-\lambda I)(x)=0 [/imath] is already zero, in which case we get the solution [imath] y=me^{\lambda t}, [/imath] or, and this is different, that [imath] (A-\lambda I)((A-\lambda I)(x))=0.[/imath] Now
[math] (A-\lambda I)(A-\lambda I)=A^2-2\lambda A +\lambda^2 I=0 [/math]and we have to solve [imath] y''-2\lambda y'+\lambda^2y=0. [/imath] Let's see if [imath] y=nte^{\lambda t} [/imath] is a solution for this equation.
[math]\begin{array}{lll}
y&=n te^{\lambda t}\\
y'&=ne^{\lambda t}+n\lambda te^{\lambda t}=n e^{\lambda t}+\lambda y\\
y''&=n\lambda e^{\lambda t}+\lambda y'=2n\lambda e^{\lambda t}+\lambda^2 y
\end{array}[/math][math]\begin{array}{lll}
y''-2\lambda y'+\lambda^2 y&=2n\lambda e^{\lambda t}+\lambda^2 y-2\lambda(n e^{\lambda t}+\lambda y)+\lambda^2 y=0
\end{array}[/math]
Hence, it is a solution. Combining both, we get
[math] y=me^{\lambda t}+n te^{\lambda t}=(m+nt)e^{\lambda t}.[/math]This is the solution space for the generalized eigenspace [imath] \lambda=\lambda_1=-3 [/imath] with multiplicity two.
If we finally add the solution space for the generalized eigenspace [imath] \lambda=\lambda_2=0 [/imath] with multiplicity one, then we get
[math] y=(m+nt)e^{\lambda_1 t}+pe^{\lambda_2 t}=(m+nt)e^{-3t}+pe^{0\cdot t}=(m+nt)e^{-3t}+p. [/math]The parameters [imath] m,n,p [/imath] result from the initial values for the differential equation.