Author Topic: TUT 0501  (Read 4286 times)

Sally

  • Newbie
  • *
  • Posts: 4
  • Karma: 0
    • View Profile
TUT 0501
« on: October 18, 2019, 02:15:33 PM »
y”+2y’+2y=0.  y(π/4)=2.      y’(π/4)=-2。        (note: π is “pie”)

r^2+2r+2=0
r1=-1+i
r2=-1-i

y=C1e^(-t)cos(t)+C2e^(-t)sin(t)

y(π/4)=2=[e^(-π/4)](1/√2)(C1+C2)
C1+C2=2/([e^(-π/4)] (1/√2))

y’(t)=-C1[e^(-t)]cos(t)-C1[e^(-t)]sin(t)
-2=-2[e^(-π/4)](1/√2)C1
C1=√2/[e^(π/4)]=C2
y=(√2/[e^(π/4)])*e^(-t)cos(t)+(√2/[e^(π/4)])*e^(-t)sin(t)