- #1
shinnsohai
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Homework Statement
Given
[itex]\frac{dx}{dt}[/itex] = -1.3x
[itex]x_{1}[/itex](t)=e[itex]^{-1.3t}[/itex]
[itex]x_{2}[/itex](t)=4e[itex]^{-1.3t}[/itex]
Compute a solution for x(t) if x(0)=3
Homework Equations
Superposition Principle
and some ODE related
Anyhow I refer to this
http://www.youtube.com/watch?v=_ECd0Jn7y68
The Attempt at a Solution
First Attempt
x(t)=[itex]\alpha[/itex] (e[itex]^{-1.3t}[/itex]) + [itex]\beta[/itex] (4e[itex]^{-1.3t}[/itex])
[itex]\frac{dx}{dt}[/itex]= (-1.3) * ([itex]\alpha[/itex] [itex]x_{1}[/itex] + [itex]\beta[/itex] [itex]x_{2}[/itex] )
Then after this step, I've no idea how to continue, I am stuck here, what Should I do with the given initial condition? x(0)=3
I've done some google search, some ODE with Initial condition provided
Perhaps Related Solution:
[itex]\frac{dy}{dx}[/itex] = -1.3x
[itex]\int[/itex] 1 dy = [itex]\int[/itex] -1.3x dx
To get the Constant, I've plugged in the given initial condition x(0)=3
y = [itex]\frac{-1.3x^{2}}{2}[/itex] + c
c = -5.58
Re-arrange the eqn
y= [itex]\frac{-1.3x^{2}}{2}[/itex] - 5.58
after getting this?
how do I proceed ?
Imma so confusee!(Sorry Mods, It's quite sometime I dint visit the forum, Making such messy mistake on the previous post)
Anyway
Here's the (Hand written working LINK)
http://imgur.com/VmfPJwr
http://imgur.com/J5k5YeO
http://imgur.com/VmfPJwr
I'm not that familiar with the Latex Code :tongue:
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