Eigenvector_Eigenvalue Method

# The bottom row of zeros gives no information so the

• Notes
• physics778899
• 6

This preview shows page 5 - 6 out of 6 pages.

= 0 0 0 . The bottom row of zeros gives no information, so the system is underdetermined and we may take v 3 to be arbitrary, say v 3 = c . The second row gives v 2 - 3 v 3 = 0 so v 2 = 3 v 3 = 3 c . The first equation gives v 1 + (3 / 2) v 3 = 0 so v 2 = - (3 / 2) v 3 = - 3 c/ 2 . Then every eigenvector corresponding to λ 1 = 5 is of the form v = - 3 c/ 2 3 c c .

Subscribe to view the full document.

Since we just need any nonzero vector of this form, we may take c = 2 to get the eigenvector v 1 = - 3 6 2 . This gives eigenvector/eigenvalue pair gives the solution y 1 = e λ 1 x v 1 = e 5 x - 3 / 2 3 1 . Repeating this same process for λ 2 = 3 we get v 2 = - 1 2 1 and hence the solution y 2 = e λ 2 x v 2 = e 3 x - 1 2 1 . And repeating again for λ 3 = 2 we get v 3 = - 1 1 1 and hence the solution y 3 = e λ 3 x v 3 = e 2 x - 1 1 1 . Since the eigenvectors were distinct, we are guaranteed that the solutions y 1 , y 2 , and y 3 are linearly independent, hence they form a basis for the solution space. Then our general solution is y ( x ) = c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 e 5 x - 3 / 2 3 1 + c 2 e 3 x - 1 2 1 + c 3 e 2 x - 1 1 1
You've reached the end of this preview.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern