Let's take the transform of every term in the differential equation
L[y′′]−L[2y′]+L[1]=0
It is known, that
L[y′′]=s2L[y]−sy(0)−y′(0)=s2Y(s)−5s+9
L[y′]=sY(s)−y(0)=sY(s)−5 .
Here Y(s)=L[y]
The Laplace-transformed differential equation is
s2Y(s)−5s+9−2(sY(s)−5)+s1=0
Y(s2−2s)=5s−s1−19
Y=s(s2−2s)5s2−19s−1=2s21+4s39−4(s−2)19
Recalling the inverse transform will lead us to
s21←x;s1←1;s−21←e2t
Therefore
Y←y=2t+439−419e2t
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