- #1

- 41

- 0

## 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, Im 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:

Last edited: