Consider the differential equation below.

$y^{}+2y^{}+0\mathrm{.}2\sqrt[\phantom{2}]{y}=0\mathrm{.}8,y(0^{-})=9,y^{}(0^{-})=0\mathrm{.}5$

$(1)$ Verify that the final value of $y(t)$ is $16.$

$(2)$ Since $y$ starts at $9$ and ends at $16,$ verify that a straight$-$line approximation for $\sqrt[\phantom{2}]{y}$ that would be

appropriate for linearizing this differential equation is $\sqrt[\phantom{2}]{y}$~~$\frac{1}{7}y+\frac{12}{7}$

$(3)$ Replace $\sqrt[\phantom{2}]{y}$ in the differential equation with your straight$-$line approximation for $\sqrt[\phantom{2}]{y}$ to get a

linear approximation for the original $2$nd order differential equation. Verify that the linear

approximation for the differential equation is:

$y^{}+2y^{}+0\mathrm{.}0286y=0\mathrm{.}457,y(0^{-})=9,y^{}(0^{-})=0\mathrm{.}5$

$(4)$ Suppose the initial value of $y$ is $16,$ not $9.$ Explain why the assumption is reasonable that $y(t)$ is

always in the neighborhood of $16.$ Since $y$ is always close to $16,$ verify that an appropriate linear

approximation for the differential equation is:

$y^{}+2y^{}+0\mathrm{.}025y=0\mathrm{.}4,y(0^{-})=16,y^{}(0^{-})=0\mathrm{.}5$