# ODE with Superposition

1. Oct 9, 2013

### shinnsohai

1. The problem statement, all variables and given/known data
Given
$\frac{dx}{dt}$ = -1.3x
$x_{1}$(t)=e$^{-1.3t}$
$x_{2}$(t)=4e$^{-1.3t}$

Compute a solution for x(t) if x(0)=3

2. Relevant equations
Superposition Principle
and some ODE related
Anyhow I refer to this

3. The attempt at a solution
First Attempt
x(t)=$\alpha$ (e$^{-1.3t}$) + $\beta$ (4e$^{-1.3t}$)
$\frac{dx}{dt}$= (-1.3) * ($\alpha$ $x_{1}$ + $\beta$ $x_{2}$ )

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:
$\frac{dy}{dx}$ = -1.3x
$\int$ 1 dy = $\int$ -1.3x dx

To get the Constant, I've plugged in the given initial condition x(0)=3
y = $\frac{-1.3x^{2}}{2}$ + c
c = -5.58

Re-arrange the eqn

y= $\frac{-1.3x^{2}}{2}$ - 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: Oct 9, 2013
2. Oct 9, 2013

### Staff: Mentor

You're making this much more difficult than it actually is. Do you know what the general solution of your differential equation is? Assuming that you do, just use your initial condition to find the solution for which x(0) = 3.
This is not at all related to your problem. A related problem would be dy/dx = -1.3y.

3. Oct 9, 2013

### shinnsohai

I've messed up the thing eventually

Based on the related problem dy/dx = -1.3 y
Integrate it
and
apply the boundary condition x(0)=3

$\int$ $\frac{dy(x)}{dx}/{y(x)}$ = $\int$ -1.3 dx

log(y(x)) = -1.3x +c

y(x) = $e^{-1.3x+c}$
Let C = $e^{C1}$

y(x) =C * $e^{-1.3x}$

Plug in the Given boundary
x(0)=3

y(x) =C * $e^{-1.3x}$
3 = C * $e^{0}$
C=3

Re-arrange

y(x) =3* $e^{-1.3x}$

So until this step
what should I do with the

$x_{1}$(t)=e$^{-1.3t}$
$x_{2}$(t)=4e$^{-1.3t}$

Last edited: Oct 9, 2013
4. Oct 9, 2013

### Staff: Mentor

Yes, that's it, although you should write your answer as x(t) = 3e-1.3t. In your diff. equation, x was the dependent variable and t was the independent variable.

I believe that what they wanted you to do with x1(t) and x2(t) was to recognize that all solutions are of the form Ce-1.3t.

5. Oct 9, 2013

### LCKurtz

You need to decide whether your independent variable is $t$ or $x$. What happens if you calculate $x_2 - x_1$?

6. Oct 9, 2013

### shinnsohai

Oh ya!
I used to use dy and dx
and now i get messed up again!:tongue:

7. Oct 9, 2013

### shinnsohai

Thanks for remind that!