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

r^{2}-2r+5=0

r_{1/2}= 1±2i

y=c_{1}e^{t}cos(2t)+c_{2}e^{t}sin(2t)

we have y(∏/2)=0, when replaced in eq above gives me

c_{1}=-e^{-(∏/2)}

calculating y' from the above y gives:

y'=e^{t}((c_{2}-2c_{1})sin(2t)+(c_{1}+2c_{2})cos(2t))

replacing y'(∏/2)=2 i get c_{2}=(3/2)e^{-(∏/2)}

putting back c_{1}and c_{2}i get final solution ofy=(3/2)e^{(t-∏/2)}sin(2t)-e^{(t-∏/2)}cos(2t)

book saying that the solution of this eq isy=-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....

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# Homework Help: 2nd order homogeneous diff eq

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