System of first order differential equations

**monster** · April 25th 2010, 08:29 PM

I'm having trouble solving this system of de's using eigenvalues and eigenvectors,

$ \frac{dx}{dt} = x - 4y $
$ \frac{dy}{dt} = x + y $
where the initial conditions are; x(0)=3 and y(0)=-4

I went through to calculate the eigenvalues and vectors to be;

$ \lambda_1 = 1 + 2i $
$ \lambda_2 = 1-2i $

and v1 = [1 , -i/2]
hence v2 = [1 , i/2]

Using this the general solution should be ?

x(t) = C1 . [1 , -i/2] $e^{(1+2i)t}$ + C2 . [1 , i/2] $e^{(1-2i)t}$

where C1 & c2 are complex constants.

have i done this right so far?
To get the actual solution with the initial conditions i'm a bit confused can anyone help me here?

Many thanks.

**Deadstar** · April 26th 2010, 05:11 AM

Originally Posted by monster

I'm having trouble solving this system of de's using eigenvalues and eigenvectors,

$ \frac{dx}{dt} = x - 4y $
$ \frac{dy}{dt} = x + y $
where the initial conditions are; x(0)=3 and y(0)=-4

I went through to calculate the eigenvalues and vectors to be;

$ \lambda_1 = 1 + 2i $
$ \lambda_2 = 1-2i $

and v1 = [1 , -i/2]
hence v2 = [1 , i/2]

Using this the general solution should be ?

x(t) = C1 . [1 , -i/2] $e^{(1+2i)t}$ + C2 . [1 , i/2] $e^{(1-2i)t}$

where C1 & c2 are complex constants.

have i done this right so far?
To get the actual solution with the initial conditions i'm a bit confused can anyone help me here?

Many thanks.

A quick calc showed me your eigenvals are correct so I'll assume the vectors are as well.

To finish this let's just take the x(0) = 3 condition first...

This says that for this equation below.

x(t) = C1 . [1 , -i/2] $e^{(1+2i)t}$ + C2 . [1 , i/2] $e^{(1-2i)t}$

When you set t=0, the 'x parts' of the LHS side should equal 3.

By 'x parts' I mean the following...

Consider that each eigenvector is of the form [ 'x', 'y'].

So you only use the 'x' parts for equating x(0) = 3.

For this example it would be...

$3 = x(0) = C_1 \cdot 1\cdot e^{(1+2i)\cdot 0} + C_2 \cdot 1 \cdot e^{(1-2i)\cdot 0}$

I.e. $3 = C_1 + C_2$

Now do the same for y(0) = -4 and you should have 2 equations with 2 unknows ( $C_1$ and $C_2$ ) and you should be able to solve it from there!

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