(adsbygoogle = window.adsbygoogle || []).push({}); 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Initial Value problem with complex roots

**Physics Forums | Science Articles, Homework Help, Discussion**