5(a)

The differential equation is $m y'' + 2 y' + y = 0$. The initial condition (taking up as positive) is $y(0) = 0.1$, $y'(0) = 0$.

With $m=2$, the characteristic polynomial $2 D^2 + 2D + 1$ has complex roots $-1/2 \pm i/2$. The general solution can be written as $y = \exp(-t/2) (c_1 \cos(t/2) + c_2 \sin(t/2))$. From the initial conditions, $y(0) = c_1 = 0.1$ and, since $y' = \exp(-t/2) ((-c_1/2 + c_2/2) \cos(t/2) - (c_1/2 + c_2/2) \sin(t/2))$, $y'(0) = -c_1/2 + c_2/2 = 0$, so $c_1 = c_2 = 0.1$ and $y = 0.1 \exp(-t/2) (\cos(t/2) + \sin(t/2))$.

Alternatively, you could write $y = A \exp(-t/2) \cos(t/2 - \phi)$ and $y' = A \exp(-t/2)(-\cos(t/2-\phi)/2 - \sin(t/2-\phi)/2)$. The initial conditions would say $y(0) = A \cos \phi= .1$ and $y'(0) = (A/2) (-\cos \phi + \sin \phi) = 0$, so $\sin \phi = \cos \phi$, $A = \sqrt{2}/10$, $\cos \phi = \sin \phi = 1/\sqrt{2}$. Thus $\phi = \pi/4$ and the solution is $y = (\sqrt{2}/10) \exp(-t/2) \cos(t/2 - \pi/4)$.