5.1

Applying the representative ODE \( u'_n = \lambda u_n+ae^{\mu hn} \)
\[ u_{n+1}=u_{n-1}+h\left[\beta_1\lambda u_n+\beta_1ae^{\mu hn}+\beta_0\lambda u_{n-1} +\beta_0ae^{\mu h(n-1)}\right] \]
and the displacement operator
\[[E-\beta_1\lambda h-(1+\beta_0\lambda h)E^{-1}]u_n = h(\beta_1+\beta_0E^{-1})ae^{\mu hn}\]
gives the characteristic polynomial
\[P(E)=E-\beta_1\lambda h-(1+\beta_0\lambda h)E^{-1} = 0\]
Solving \( P(\sigma) = 0 \) gives
\[\sigma_{1,2}=\frac{1}{2}\beta_1\lambda h \pm \frac{1}{2}\sqrt{\beta_1^2\lambda^2h^2+4\beta_0\lambda h+4}\]
Rearranging and using the binomial expansion for the square root gives
\[\sigma_1=\frac{1}{2}\beta_1\lambda h+\sqrt{1+\left(\beta_0\lambda h+\frac{1}{4}\beta_1^2\lambda^2h^2\right)}=1+\frac{1}{2}\beta_1\lambda h+\frac{1}{2}\left(\beta_0\lambda h+\frac{1}{4}\beta_1^2\lambda^2h^2\right)-\frac{1}{8}\left(\beta_0\lambda h+\frac{1}{4}\beta_1^2\lambda^2h^2\right)^2\]
For the exact solution, \( \sigma = e^{\lambda h} = 1 + \lambda h + \frac{1}{2}\lambda^2h^2 + \cdots \)

For first-order accuracy,
\[\sigma_1 = 1+\lambda h+\cdots = 1+\frac{1}{2}\left(\beta_1+\beta_0\right)\lambda h+\cdots \]
equating the \( \lambda h \)-term gives the constraint \( \beta_1 + \beta_0 = 2 \).

For second-order accuracy,
\[\sigma_1 = 1+\lambda h+\frac{1}{2}\lambda^2h^2+\cdots = 1+\frac{1}{2}\left(\beta_1+\beta_0\right)\lambda h+\frac{1}{8}\left(\beta_1^2-\beta_0^2\right)\lambda^2h^2+\cdots \]
equating the \( \lambda h \) and \( \lambda^2 h^2 \)-terms gives the constraints \( \beta_0 = 0 \) and \( \beta_1 = 2 \). This method is called ``Leap-Frog".