Author Topic: TUT0602 Quiz 1  (Read 505 times)

Yuchen Cong

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TUT0602 Quiz 1
« on: September 27, 2019, 04:05:25 PM »
Q: y'-2y=e2t, y(0)=2

Solution:

p(t)=-2, g(t)=e2t
μ(t)=e∫ -2 dt=e-2t

Multiply both sides by μ(t):
e-2ty'-2e-2ty=e2t*e-2t
(e-2ty)'=e0=1

Integrate both sides:
∫(e-2ty)'=∫ 1 dt
e-2ty=t+C, where C is a constant

Divide both sides by e-2t to isolate y:
y=(t+C)/e-2t
y=(t+C)e2t

Substituting y(0)=2:
2=(0+C)e0
2=C

Thus,
y=(t+2)e2t[/pre]