The bottom row of zeros gives no information so the

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= 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 .
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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
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