Question

2nd order ODE problem.

Answer 1

Trying to solve the IVP the last part of the question.

Answer 2

I've found the complex solutions and the real solutions

Answer 3

just plug in the initial values into the solutions to define a specific result

prolly have to do the x' first in order to establish any Constants that need defined

Answer 4

I have find the derivative of the complex solution first though.

Answer 5

what is your x(t)?

Answer 6

This was the complex solution I got from the first part

Answer 7

x(t) is not y(x) .... are you simply ignoring the naming conventions? or is this the result of some other problem?

Answer 8

you are going to have to solve a system of equations in this case

Answer 9

if you are using latex for this, please write your results in here so that we dont have to play badmiton going back and forth to verify things

Answer 10

Ok what I did for b(i) was get the following complex root after using the quadratic formula \[(-6 \pm \sqrt{-36})/4\]

Answer 11

\[[e^u(c_1cosu+sinu)]'=e'^u(c_1cosu+sinu)+e^u(c_1cosu+sinu)'\]
\[[e^u(c_1cosu+sinu)]'=u'e^u(c_1cosu+sinu)+e^u(-u'c_1sinu+u'cosu)\]

Answer 12

Simplifies down to\[-1.5 + 1.5i \] and \[ -1.5 - 1.5i\]

Answer 13

im not sure why you are using complex roots to find x'

Answer 14

I'm based my answer based on this, it's the same question except the girl in the video has a distinct roots and not complex roots. http://www.youtube.com/watch?v=uM9h47YFE7o

Answer 15

You get your solution, and the deriviate of solution sub in your initial conditions and you're will be left with a simultaneous equation.

Answer 16

your making this way to difficult on yourself

Answer 17

\[y(x) = c_1e^{-1.5x}cos(1.5x)+c_2e^{-1.5x}sin(1.5x)\]
\[y'(x) = c_1[-(1.5)e^{-1.5x}cos(1.5x)-(1.5)e^{-1.5x}sin(1.5x)]\]\[\hspace{3em}+c_2[-(1.5)e^{-1.5x}sin(1.5x)+(1.5)e^{-1.5x}cos(1.5x)]\] now fill in your initial values
0,0 and 0,3

\[0 = c_1e^{0}cos(0)+c_2e^{0}sin(0)\]
\[3 = c_1[-(1.5)e^{0}cos(0)-(1.5)e^{0}sin(0)]\]\[\hspace{3em}+c_2[-(1.5)e^{0}sin(0)+(1.5)e^{0}cos(0)]\]

\[0 = c_1\]\[3 = c_1[-(1.5)]+c_2[(1.5)]\]

c1 = 0, and c2 = 3/1.5

Answer 18

which gives us\[x = 2e^{-1.5t}sin(1.5t)\]

Answer 19

Are you using the product rule there?

Answer 20

yes

Answer 21

@amistre64 Thank you, the illustrious chain rule..screwing me over once again. I didn't see it. If you ever come to Ireland, I owe you a pint for all the ODE help!

Answer 22

youre welcome :)