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**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

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 ??

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.