Q: y'-2y=e^{2t}, y(0)=2
Solution:
p(t)=-2, g(t)=e^{2t}
μ(t)=e^{∫ -2 dt}=e^{-2t}
Multiply both sides by μ(t):
e^{-2t}y'-2e^{-2t}y=e^{2t}*e^{-2t}
(e^{-2t}y)'=e^{0}=1
Integrate both sides:
∫(e^{-2t}y)'=∫ 1 dt
e^{-2t}y=t+C, where C is a constant
Divide both sides by e^{-2t} to isolate y:
y=(t+C)/e^{-2t}
y=(t+C)e^{2t}
Substituting y(0)=2:
2=(0+C)e^{0}
2=C
Thus,
y=(t+2)e^{2t}[/pre]