# Some Diffeq / Matrices help

• Jan 25th 2009, 08:06 PM
Codename46
Some Diffeq / Matrices help
Posted this in urgent to no avail:

I'm having some slight trouble solving some Applied LinAlg problems, namely Decoupling Equations.

Suppose you have a matrix A with the contents

( 2 0 )
( 0 3 )

I need to write down the solution to the equation dx/dt = Ax (A being the matrix).

I need to express x1(t) and x2(t) (x sub 1, x sub 2) in terms of initial conditions x1(0) and x2(0).

So far I've got

dx1/dt = 2x1
dx2/dt = 3x2

I'm not sure how to put the initial conditions in here and solve it.
• Jan 25th 2009, 08:31 PM
mr fantastic
Quote:

Originally Posted by Codename46
Posted this in urgent to no avail:

I'm having some slight trouble solving some Applied LinAlg problems, namely Decoupling Equations.

Suppose you have a matrix A with the contents

( 2 0 )
( 0 3 )

I need to write down the solution to the equation dx/dt = Ax (A being the matrix).

I need to express x1(t) and x2(t) (x sub 1, x sub 2) in terms of initial conditions x1(0) and x2(0).

So far I've got

dx1/dt = 2x1
dx2/dt = 3x2

I'm not sure how to put the initial conditions in here and solve it.

Why can't you just solve each DE seperately using the usual techniques ....?
• Jan 25th 2009, 08:46 PM
Codename46
Quote:

Originally Posted by mr fantastic
Why can't you just solve each DE seperately using the usual techniques ....?

Err, haha my bad. Apparently the diagonal matrix makes each DE uncoupled instead of coupled.

In any case, I'm not sure what they mean by "in terms of initial conditions", and my diffeq skills are rusty. How do I incorporate the initial conditions to get the solution?
• Jan 25th 2009, 09:16 PM
mr fantastic
Quote:

Originally Posted by Codename46
Err, haha my bad. Apparently the diagonal matrix makes each DE uncoupled instead of coupled.

In any case, I'm not sure what they mean by "in terms of initial conditions", and my diffeq skills are rusty. How do I incorporate the initial conditions to get the solution?

What are the initial conditions? All you say is x1(0) and x2(0).

You should know how to solve a differential equation that has the simple form that these two have. If you don't, then it's your job to go back and review the necessary material in your class notes or textbook or on-line (using Google).

If you know the value of x when t = 0, then you substitute that information into your general solution to solve for the arbitrary constant.
• Jan 25th 2009, 09:17 PM
Codename46
Quote:

Originally Posted by mr fantastic
What are the initial conditions? All you say is x1(0) and x2(0).

You should know how to solve a differential equation that has the simple form that these two have. If you don't, then it's your job to go back and review the necessary material in your class notes or textbook or on-line (using Google).

If you know the value of x when t = 0, then you substitute that information into your general solution to solve for the arbitrary constant.

They don't define x1(0) and x2(0), so I'm assuming there's nothing to substitute.
• Jan 25th 2009, 09:28 PM
mr fantastic
Quote:

Originally Posted by Codename46
They don't define x1(0) and x2(0), so I'm assuming there's nothing to substitute.

Then the best you can do is something like $x_1 = x_1(0)$ when $t = 0$ etc. and substitute that.