Distinct eigenvalues –system solving, Mathematics

DISTINCT EIGENVALUES -SYSTEM SOLVING:

Example

Solve the following IVP.

1197_DISTINCT EIGENVALUES –SYSTEM SOLVING.png

Solution:

Therefore, the first thing that we must to do that is, get the eigenvalues for the matrix.

1334_DISTINCT EIGENVALUES –SYSTEM SOLVING1.png

= l2 -3l - 4

= (l+ 1) (l- 4)                                     ⇒                     l1 = -1,  l2 = 4

Here let's get the eigenvectors for each of these.

For l1 = -1,

We have to solve:

2433_DISTINCT EIGENVALUES –SYSTEM SOLVING2.png

= 2h1 + 2h2                              ⇒                                 h1 = - h2

The eigenvector for this case is,

2482_DISTINCT EIGENVALUES –SYSTEM SOLVING3.png

= h2= 1,

For l2 = 4,

We have to solve:

1479_DISTINCT EIGENVALUES –SYSTEM SOLVING4.png

⇒ = -3h1 + 2h2 = 0                              ⇒                                 h1 =( 2/3) h2

The eigenvector for this case is,

1101_DISTINCT EIGENVALUES –SYSTEM SOLVING5.png

= h2 = 3

So the general solution is after that,

1221_DISTINCT EIGENVALUES –SYSTEM SOLVING6.png

Now, we have to get the constants. To do that we simply have to apply the initial conditions

530_DISTINCT EIGENVALUES –SYSTEM SOLVING7.png

All we required to do now is multiply the constants with and we after that get two equations that is one for each row which we can solve for the constants. It gives,

-c1 + 2c2 = 0;

c1 + 3c2 = -4;

By solving both equations we get:

c1 = -(8/5) and c2 = (-4/5)

The solution is subsequently,

 

1221_DISTINCT EIGENVALUES –SYSTEM SOLVING8.png

Posted Date: 4/11/2013 1:30:55 AM | Location : United States







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