2nd order homogeneous diff eq

1. Oct 6, 2012

leonida

1. The problem statement, all variables and given/known data
y"-2y'+5=0, y(∏/2)=0, y'(∏/2)=2

find general solution of this diff eq

2. Relevant equations

3. The attempt at a solution

i have followed all of the steps for this, rather easy 2nd order diff eq, but i my solution differs from the books solution.

steps:
r2-2r+5=0

r 1/2= 1±2i

y=c1etcos(2t)+c2etsin(2t)
we have y(∏/2)=0, when replaced in eq above gives me
c1=-e-(∏/2)

calculating y' from the above y gives:
y'=et((c2-2c1)sin(2t)+(c1+2c2)cos(2t))
replacing y'(∏/2)=2 i get c2=(3/2)e-(∏/2)

putting back c1 and c2 i get final solution of y=(3/2)e(t-∏/2)sin(2t)-e(t-∏/2)cos(2t)
book saying that the solution of this eq is y=-e(t-∏/2)sin(2t)

can someone point out where i am making mistakes, since i did this problem few times and i always get the same answer....

2. Oct 6, 2012

vela

Staff Emeritus
This actually gives you $y(\pi/2)=0=c_1e^{\pi/2}$. You set it equal to 1, not 0.

3. Oct 6, 2012

leonida

vela,
thank you.. i knew i made some silly mistake...

after correcting my work i still have discrepancy in sign.. i am getting positiv, while book is saying negative..

After differentiating y I get => y'=et((c2-2c1)sin(2t)+(c1+2c2)cos(2t))

since c1=0 and y'(∏/2)=2

2=2e∏/2c2 which is

c2=e-(∏/2) and i need this to be negative......

4. Oct 6, 2012

leonida

got it.. cos(pi)=-1 and there i get my negative sign... Vela, thanks alot for pointing my mistake

5. Oct 6, 2012

vela

Staff Emeritus