Example of repeated eigenvalues, Mathematics

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Illustration: Solve the following IVP.

1016_Example of Repeated eigenvalues.png

Solution:

First get the eigenvalues for the system.

1604_Example of Repeated eigenvalues1.png

= l2 - 10 l+ 25

= (l- 5)2

l1,2 = 5

Therefore, we got a double eigenvalue. Obviously that must not be too surprising given the section which we're in. here we find the eigenvector for that eigenvalue.

704_Example of Repeated eigenvalues2.png

2h1 +   h2 = 0,                         ⇒         h2 = - 2h1

446_Example of Repeated eigenvalues3.png

The eigenvector is,

h1≠ 0

h1= 1

The next step is get ?r.  To do this we'll require solving,

848_Example of Repeated eigenvalues4.png

2  ?r1+  ?r2 = 1,                       ?r2 = 1 - 2  ?r1

Remember that this is almost the same to the system which we solve to find the eigenvalue.  The simple difference is the right hand side. The most common possible ?r is,

601_Example of Repeated eigenvalues5.png

If r1 = 0

During this case, unlike the eigenvector system we can select the constant to be anything we need, therefore we might as well pick it to create our life easier. This generally means picking this to be zero.

 We can now write down the general solution to the system.

109_Example of Repeated eigenvalues6.png

Applying the initial condition to get the constants provides us,

2087_Example of Repeated eigenvalues7.png

c1 = 2;

-2 c1 + c2 = -5;

By solving both equations we get:

c1 = 2;

c2 = -1

The actual solution is,

1353_Example of Repeated eigenvalues8.png


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