2nd order homogeneous diff eq

[leonida]

Homework Statement

y"-2y'+5=0, y(∏/2)=0, y'(∏/2)=2

find general solution of this diff eq

Homework Equations

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=c₁e^tcos(2t)+c₂e^tsin(2t)
we have y(∏/2)=0, when replaced in eq above gives me
c₁=-e^-(∏/2)

calculating y' from the above y gives:
y'=e^t((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...

 

[vela]

Homework Statement

y"-2y'+5=0, y(∏/2)=0, y'(∏/2)=2

find general solution of this diff eq

Homework Equations

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=c₁e^tcos(2t)+c₂e^tsin(2t)
we have y(∏/2)=0, when replaced in eq above gives me
c₁=-e^-(∏/2)

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

calculating y' from the above y gives:
y'=e^t((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...

 

[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'=e^t((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...

 

[leonida]

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

 

[vela]

Glad you figured it out.