Quantum Chemistry

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Exercise. This is the solution to exercise 2.1 in the book.

Solution.

(a) The characteristic equation is:

m2 4m + 3 = 0

m1 = 1m2 = 3

Thus the general solution is:

y(x) = Aex + Be3x

(b) The characteristic equation is:

m2 + 6m = 0

m1 = 0m2 = 6

Thus the general solution is:

y(x) = A + Be6x

(c) We can write

dy y = 3dx

lny = 3x + C

or

y(x) = Ae3x

(d) The characteristic equation is:

m2 + 2m 1 = 0

m1 = 1 + 2m2 = 1 2

Thus the general solution is:

y(x) = ex Ae2x + Be2x

(e) The characteristic equation is:

m2 3m + 2 = 0

m1 = 1m2 = 2

Thus the general solution is:

y(x) = Aex + Be2x