How Can the Superposition Principle Help Solve an ODE with Initial Conditions?

[shinnsohai]

Homework Statement

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

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)=\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, 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:
\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

 

[Mark44]

Homework Statement

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

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.

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

This is not at all related to your problem. A related problem would be dy/dx = -1.3y.

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

 

[shinnsohai]

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.

Thanks Mark for replying

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}

 

[Mark44]

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.

 

[LCKurtz]

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}

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

 

[shinnsohai]

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.

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

 

[shinnsohai]

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

Thanks for remind that!