Quantum Chemistry

Chapter02 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29

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

Solution. The characteristic equation is:

m2 + m 6 = 0

with solutions

m1 = 2m2 = 3

Hence, the general solution is:

y(x) = c1e2x + c 2e3x

For

y(0) = 0 y(0) = 1

or

0 = c1 + c2 1 = 2c1 3c2

Hence:

y(x) = 1 5e2x 1 5e3x