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

Answer:

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.

 

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).

 

hey guys. i finally figured out why the answer was

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

this answer is correct, and my initial answer was incorrect.

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.

 

to be in save side i use Maple program

 

it got to be easy !