- #1
ManMonkeyFish
- 5
- 0
1.Solve the following initial value problem
d^2y/dt^2 + 4y = { 1, 0<= t <=Pi/2 } {sin(t), Pi/2 < t}
y(0) = 2
y'(0) = -1
I used y(t) = e^rt
3. For y'' + 4y' =1
y(t)= e^rt
Yp(0)=1
r^2+4 = 0
r = 2i
y1(t) = C1 Cos(2t) + C2Sin(2t) + Yp(0)
y1'(t) = -2C1Sin(2t) + 2C2Cos(2t)
y1(0) = 2 = C1 +1
y1'(0) = -1 = 2C2
C1 = 1
C2 = -1/2
y1(t) = Cos(2t) - 1/2 Sin(2t) + 1
y1'(t) = -2Sin(2t) - Cos(2t)
y1(Pi/2) = 0 ??
y1'(Pi/2) = 1
for t > Pi/2
0 = y1(Pi/2) = C1
1 = y1'(pi/2) = 2C2
C1 = 0
C2 = 1/2
so my answer is
-2Sin(2t) - Cos(2t) 0<= t <= Pi/2
Cos(2t) Pi/2 < t
Which I believe is wrong, I think I made a mistake at this step
y1(t) = Cos(2t) - 1/2 Sin(2t) + 1
y1(Pi/2) = 0 ??
Thank you for any help, this one has been giving me problems:yuck: .
I need to find out where I am going wrong, and I appreciate any and all feedback.
d^2y/dt^2 + 4y = { 1, 0<= t <=Pi/2 } {sin(t), Pi/2 < t}
y(0) = 2
y'(0) = -1
I used y(t) = e^rt
3. For y'' + 4y' =1
y(t)= e^rt
Yp(0)=1
r^2+4 = 0
r = 2i
y1(t) = C1 Cos(2t) + C2Sin(2t) + Yp(0)
y1'(t) = -2C1Sin(2t) + 2C2Cos(2t)
y1(0) = 2 = C1 +1
y1'(0) = -1 = 2C2
C1 = 1
C2 = -1/2
y1(t) = Cos(2t) - 1/2 Sin(2t) + 1
y1'(t) = -2Sin(2t) - Cos(2t)
y1(Pi/2) = 0 ??
y1'(Pi/2) = 1
for t > Pi/2
0 = y1(Pi/2) = C1
1 = y1'(pi/2) = 2C2
C1 = 0
C2 = 1/2
so my answer is
-2Sin(2t) - Cos(2t) 0<= t <= Pi/2
Cos(2t) Pi/2 < t
Which I believe is wrong, I think I made a mistake at this step
y1(t) = Cos(2t) - 1/2 Sin(2t) + 1
y1(Pi/2) = 0 ??
Thank you for any help, this one has been giving me problems:yuck: .
I need to find out where I am going wrong, and I appreciate any and all feedback.