1. The mass [latex]m=0.0427\text{kg},k=0.5[/latex], and [latex]g=9.8{\text{m/s}}^{2}[/latex]. The initial velocity is [latex]{v}_{0}=2[/latex] m/s. Therefore the initial-value problem is
   
   [latex]0.0427\frac{dv}{dt}=-0.5v - 0.0427\left(9.8\right),{v}_{0}=2[/latex].

   
   
   Dividing the differential equation by [latex]0.0427[/latex] gives
   

   [latex]\frac{dv}{dt}=-11.7096v - 9.8,{v}_{0}=2[/latex].

    

   
   
   The differential equation is linear. Using the problem-solving strategy for linear differential equations:
   
   Step 1. Rewrite the differential equation as [latex]\frac{dv}{dt}+11.7096v=-9.8[/latex]. This gives [latex]p\left(t\right)=11.7096[/latex] and [latex]q\left(t\right)=-9.8[/latex]
   
   Step 2. The integrating factor is [latex]\mu \left(t\right)={e}^{\displaystyle\int 11.7096dt}={e}^{11.7096t}[/latex].
   
   Step 3. Multiply the differential equation by [latex]\mu \left(t\right)\text{:}[/latex]
   

   [latex]\begin{array}{ccc}\hfill {e}^{11.7096t}\frac{dv}{dt}+11.7096v{e}^{11.7096t}& =\hfill & -9.8{e}^{11.7096t}\hfill \\ \hfill \frac{d}{dt}\left[v{e}^{11.7096t}\right]& =\hfill & -9.8{e}^{11.7096t}.\hfill \end{array}[/latex]

   
   
   Step 4. Integrate both sides:
   

   [latex]\begin{array}{ccc}\hfill {\displaystyle\int \frac{d}{dt}\left[v{e}^{11.7096t}\right]dt}& =\hfill & {\displaystyle\int -9.8{e}^{11.7096t}dt}\hfill \\ \hfill v{e}^{11.7096t}& =\hfill & \frac{-9.8}{11.7096}{e}^{11.7096t}+C\hfill \\ \hfill v\left(t\right)& =\hfill & -0.8369+C{e}^{-11.7096t}.\hfill \end{array}[/latex]

   
   
   Step 5. Solve for [latex]C[/latex] using the initial condition [latex]{v}_{0}=v\left(0\right)=2\text{:}[/latex]
   

   [latex]\begin{array}{ccc}\hfill v\left(t\right)& =\hfill & -0.8369+C{e}^{-11.7096t}\hfill \\ \hfill v\left(0\right)& =\hfill & -0.8369+C{e}^{-11.7096\left(0\right)}\hfill \\ \hfill 2& =\hfill & -0.8369+C\hfill \\ \hfill C& =\hfill & 2.8369.\hfill \end{array}[/latex]

   
   
   Therefore the solution to the initial-value problem is [latex]v\left(t\right)=2.8369{e}^{-11.7096t}-0.8369[/latex].

2. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:
   
   [latex]\begin{array}{}\\ \hfill 2.8369{e}^{-11.7096t}-0.8369& =\hfill & 0\hfill \\ \hfill 2.8369{e}^{-11.7096t}& =\hfill & 0.8369\hfill \\ \hfill {e}^{-11.7096t}& =\hfill & \frac{0.8369}{2.8369}\approx 0.295\hfill \\ \hfill \text{ln}{e}^{-11.7096t}& =\hfill & \text{ln}0.295\approx -1.221\hfill \\ \hfill -11.7096t& =\hfill & -1.221\hfill \\ \hfill t& \approx \hfill & 0.104.\hfill \end{array}[/latex]

   
   
   Therefore it takes approximately [latex]0.104[/latex] second to reach maximum height.

3. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if [latex]h\left(t\right)[/latex] represents the height at time [latex]t[/latex], then [latex]{h}^{\prime }\left(t\right)=v\left(t\right)[/latex]. Because we know [latex]v\left(t\right)[/latex] and the initial height, we can form an initial-value problem:
   
   [latex]{h}^{\prime }\left(t\right)=2.8369{e}^{-11.7096t}-0.8369,h\left(0\right)=1[/latex].

   
   
   Integrating both sides of the differential equation with respect to [latex]t[/latex] gives
   

   [latex]\begin{array}{ccc}\hfill {\displaystyle\int {h}^{\prime }\left(t\right)dt}& =\hfill & {\displaystyle\int 2.8369{e}^{-11.7096t}-0.8369dt}\hfill \\ \hfill h\left(t\right)& =\hfill & -\frac{2.8369}{11.7096}{e}^{-11.7096t}-0.8369t+C\hfill \\ \hfill h\left(t\right)& =\hfill & -0.2423{e}^{-11.7096t}-0.8369t+C.\hfill \end{array}[/latex]

   
   
   Solve for [latex]C[/latex] by using the initial condition:
   

   [latex]\begin{array}{ccc}\hfill h\left(t\right)& =\hfill & -0.2423{e}^{-11.7096t}-0.8369t+C\hfill \\ \hfill h\left(0\right)& =\hfill & -0.2423{e}^{-11.7096\left(0\right)}-0.8369\left(0\right)+C\hfill \\ \hfill 1& =\hfill & -0.2423+C\hfill \\ \hfill C& =\hfill & 1.2423.\hfill \end{array}[/latex]

   
   
   Therefore
   

   [latex]h\left(t\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423[/latex].

   
   
   After [latex]0.104[/latex] second, the height is given by
   
   [latex]h\left(0.2\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423\approx 1.0836[/latex] meter.

We have a resistor and an inductor in the circuit, so we use [latex]L{i}^{\prime }+Ri+\frac{1}{C}q=E[/latex]. The voltage drop across the resistor is given by [latex]{E}_{R}=Ri=5i[/latex]. The voltage drop across the inductor is given by [latex]{E}_{L}=L{i}^{\prime }=0.4{i}^{\prime }[/latex]. The electromotive force becomes the right-hand side of the equation. Therefore the equation becomes

Dividing both sides by [latex]0.4[/latex] gives the equation

Since the initial current is 0, this result gives an initial condition of [latex]i\left(0\right)=0[/latex]. We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as [latex]{i}^{\prime }+12.5i=125\sin{20t}[/latex]. This gives [latex]p\left(t\right)=12.5[/latex] and [latex]q\left(t\right)=125\sin20t[/latex].

Step 2. The integrating factor is [latex]\mu \left(t\right)={e}^{\displaystyle\int 12.5dt}={e}^{12.5t}[/latex].

Step 3. Multiply the differential equation by [latex]\mu \left(t\right)\text{:}[/latex]

Step 4. Integrate both sides:

Step 5. Solve for [latex]C[/latex] using the initial condition [latex]v\left(0\right)=2\text{:}[/latex]

Therefore the solution to the initial-value problem is [latex]i\left(t\right)=\frac{250\sin{20t} - 400\cos{20t}+400{e}^{-12.5t}}{89}=\frac{250\sin{20t} - 400\cos{20t}}{89}+\frac{400{e}^{-12.5t}}{89}[/latex].

The first term can be rewritten as a single cosine function. First, multiply and divide by [latex]\sqrt{{250}^{2}+{400}^{2}}=50\sqrt{89}\text{:}[/latex]

Next, define [latex]\phi [/latex] to be an acute angle such that [latex]\cos\phi =\frac{8}{\sqrt{89}}[/latex]. Then [latex]\sin\phi =\frac{5}{\sqrt{89}}[/latex] and

Therefore the solution can be written as

The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by [latex]\phi [/latex], and the amplitude of the steady-state current is given by [latex]\frac{50\sqrt{89}}{89}[/latex]. The graph of this solution appears in Figure 3:

Figure 3.