Analytical solution of these coupled differential equations

[gursimran]

Homework Statement

I don't know how to type math equations of I have included a image file. Take initial conditiona as [0 1]

Homework Equations

The Attempt at a Solution

No idea

 

[micromass]

"No idea" isn't really an attempt...

What do you know of differential equations? Did you already solve things like this?

 

[HallsofIvy]

This is your homework problem- obviously your teacher thinks you should know how to do it yourself! Another problem is there there are several different ways to do this (finding eigenvalues and eigenvectors of a matrix or reducing to a single second order equation, or ...) and we have no idea which you have been taught.

 

[gursimran]

My teacher dinnt give me this. I figured out this on internet. I'm bad in differential equations. I thought these coupled diff equations can be solved simultaneously and then integration but no such luck cos of one being exactly opp of other ..

 

[micromass]

My teacher dinnt give me this. I figured out this on internet. I'm bad in differential equations. I thought these coupled diff equations can be solved simultaneously and then integration but no such luck cos of one being exactly opp of other ..

First read http://en.wikipedia.org/wiki/Matrix_exponential then, the interesting part is the section of "applications" which solves a problem very much like yours!

 

[hunt_mat]

Or failing matrix exponentiation, you can decouple them; Differentiate the top equation to find:
[tex]
\frac{d^{2}A}{dt^{2}}=-10\frac{dA}{dt}+50\frac{dB}{dt}
[/tex]
Now use the second equation to substitute for dB/dt, this will leave you with a B in your equation which can be gotten rid of by using the first equation. This will leave you with a second order differential equation for A.

 

[gursimran]

Or failing matrix exponentiation, you can decouple them; Differentiate the top equation to find:
[tex]
\frac{d^{2}A}{dt^{2}}=-10\frac{dA}{dt}+50\frac{dB}{dt}
[/tex]
Now use the second equation to substitute for dB/dt, this will leave you with a B in your equation which can be gotten rid of by using the first equation. This will leave you with a second order differential equation for A.

oh thanks a lot, this is a much better and simple solution.