TUT0602 Quiz 1

[Yuchen Cong]

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^-2ty'-2e^-2ty=e^2t*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)e^2t

Substituting y(0)=2:
2=(0+C)e⁰
2=C

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