# 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

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

(a) The characteristic equation is:

${m}^{2}-4m+3=0$

${m}_{1}=1\phantom{\rule{1em}{0ex}}{m}_{2}=3$

Thus the general solution is:

$y\left(x\right)=A{e}^{x}+B{e}^{3x}$

(b) The characteristic equation is:

${m}^{2}+6m=0$

${m}_{1}=0\phantom{\rule{1em}{0ex}}{m}_{2}=-6$

Thus the general solution is:

$y\left(x\right)=A+B{e}^{-6x}$

(c) We can write

$\frac{dy}{y}=-3dx$

$lny=-3x+\mathsc{C}$

or

$y\left(x\right)=A{e}^{-3x}$

(d) The characteristic equation is:

${m}^{2}+2m-1=0$

${m}_{1}=-1+\sqrt{2}\phantom{\rule{1em}{0ex}}{m}_{2}=-1-\sqrt{2}$

Thus the general solution is:

$y\left(x\right)={e}^{-x}\left(A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}\right)$

(e) The characteristic equation is:

${m}^{2}-3m+2=0$

${m}_{1}=1\phantom{\rule{1em}{0ex}}{m}_{2}=2$

Thus the general solution is:

$y\left(x\right)=A{e}^{x}+B{e}^{2x}$