# 2nd ODE with initial conditions

Discussion in 'Math' started by squirby, Aug 24, 2009.

1. ### squirby Thread Starter Member

Aug 21, 2009
15
0
hey guys.

wondering if anyone can explain how they got to this stage:

y''(t) + 2y'(t) + 5y(t) = 0 , Initial conditions: y(0) = 2, y'(0) = -4

y(t) = [√5 exp(-t) cos{ 2t + pi - arctan(1/2)]

when i work it out, i get

y(t) = exp(-t) [2cos2t - 2sin2t]

any help is much appreciated.

2. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Your answer looks correct to me, and the other answer looks incorrect. It's very easy to check whether your answer is correct, though. If it satisfies the initial conditions and differential equation, it is correct. If the other answer doesn't satisfy the initial conditions and differential equation, it is correct.

For the differential equation, the solution must be $y = c_1e^{-t}cos(2t)+ c_2e^{-t}sin(2t)$
for some constants $c_1~and~c_2$.
The constants are determined by the initial conditions y(0) and y'(0).

3. ### squirby Thread Starter Member

Aug 21, 2009
15
0
hey guys. i finally figured out why the answer was

y(t) = [√5 exp(-t) cos{ 2t + pi - arctan(1/2)].

the correct answer should have been

y(t) = exp(-t) [2cos2t - sin2t]. while this is correct, it can also be expressed in the form above.

as you know, when an solution to a differential equation has imaginary roots, it can be expressed as:

y(t) = exp(At)[CcosMt + D sinMt] where the roots of the solution to the differential equation is A ± Mj where j is the imaginary number √-1 and A and M are real numbers.

however, y(t) can also be expressed in another fashion:

y(t) = [√(C^2 + D^2)]exp(At)cos[Mt - arctan (D/C)].

applying this to y(t) = exp(-t) [2cos2t - sin2t] you arrive at y(t) = [√5 exp(-t) cos{ 2t + pi - arctan(1/2)]. this is a rule apparently, although I have never seen it in any textbook. my electrical tutor told me about it.

anyways just thought i'd share the info wif u guys.

4. ### volume New Member

Sep 7, 2009
4
0
to be in save side i use Maple program

5. ### ][ Shocked ][ Member

Apr 13, 2009
22
0
it got to be easy !