Linear Second-Order Equations

Solve the initial value problem:

y'" - 2y" - y'+ 2y = 0

y(0) = 2,   y'(0) = 3,   y"(0) = 5

By using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation, find general solutions to the following equations.

$$\begin{gathered} \left(\mathit{a}\right) \mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{18}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{19}\mathit{y}\mathbf{=}\mathbf{0} \\ \left(\mathit{b}\right) {\mathit{y}}^{\mathit{i}\mathit{v}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0} \\ \left(\mathit{c}\right) {\mathit{y}}^{\mathit{v}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathit{i}\mathit{v}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{15}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

One way to define hyperbolic functions is by means of differential equations. Consider the equation $\mathit{y}\text{'}\text{'}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$. The hyperbolic cosine, $\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$, is defined as the solution of this equation subject to the initial values: $\mathit{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$ and $\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. The hyperbolic sine, $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$, is defined as the solution of this equation subject to the initial values: $\mathit{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$ and $\mathit{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

1. Solve these initial value problems to derive explicit formulas for $\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$ and $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$. Also, show that $\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$ and $\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$.
2. Prove that a general solution of the equation $\mathit{y}\text{'}\text{'}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$ is given by $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$.
3. Suppose a,b and c are given constants for which $\mathit{a}{\mathit{r}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{r}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$ has two distinct real roots. If the two roots are expressed in the form $\mathit{\alpha }\mathbf{-}\mathit{\beta }$ and $\mathit{\alpha }\mathbf{+}\mathit{\beta }$, show that a general solution of the equation $\mathit{a}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{b}\mathit{y}\text{'}\mathbf{+}\mathit{c}\mathit{y}\mathbf{=}\mathbf{0}$ is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathit{\alpha }\mathit{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{\left(}\mathit{\beta }\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathit{\alpha }\mathit{t}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{\left(}\mathit{\beta }\mathit{t}\mathbf{\right)}$.
4. Use the result of the part (c) to solve the initial value problem: $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\text{'}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{17}\mathbf{/}\mathbf{2}$.

The auxiliary equation for the given differential equation has complex roots. Find a general solution $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}$. 

The auxiliary equation for the given differential equation has complex roots. Find a general solution $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$. 

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{z}\mathbf{=}\mathbf{0}$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{w}\mathbf{=}\mathbf{0}$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the given differential equation has complex roots. Find a general solution $\mathbf{4}\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{y}\text{'}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$. 

Find a general solution. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{4}\mathbf{y}\text{'}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{z}\mathbf{"}\mathbf{+}\mathbf{10}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{z}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{u}\text{'}\text{'}\mathbf{+}\mathbf{7}\mathbf{u}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution. $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{3}\mathbf{y}\text{'}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{10}\mathbf{y}\text{'}\mathbf{+}\mathbf{41}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{-}y\text{'}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution. $\mathbf{2}\mathbf{y}"\mathbf{+}\mathbf{13}\mathbf{y}\text{'}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution $\mathbf{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{3}\mathbf{y}\text{'}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Find a general solution.$\mathbf{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{c}\mathit{s}{\mathit{c}}^{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$

Solve the given initial value problem $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{17}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.    

Solve the given initial value problem. $\mathbf{w}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{w}\text{'}\mathbf{+}\mathbf{2}\mathbf{w}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{w}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}$$\mathbf{w}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1},$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$  

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\pi }\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{7}\mathbf{y}\text{'}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}1\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$$\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

To see the effect of changing the parameter in the initial value problem. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{b}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Solve the problem for b=5, 4 and 2 and sketch the solutions.

Find a general solution to the following higher-order equations.

(a) $\mathit{y}\text{'}\text{'}\text{'}\mathbf{-}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{y}\text{'}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

(b) $\mathit{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathit{y}\text{'}\text{'}\mathbf{+}\mathbf{5}\mathit{y}\text{'}\mathbf{-}\mathbf{26}\mathit{y}\mathbf{=}\mathbf{0}$

(c) ${\mathit{y}}^{iv}\mathbf{+}\mathbf{13}\mathit{y}\text{'}\text{'}\mathbf{+}\mathbf{36}\mathit{y}\mathbf{=}\mathbf{0}$

Using the representation for ${\mathit{e}}^{\mathbf{(}\mathit{\alpha }\mathbf{+}\mathit{i}\mathit{\beta }\mathbf{)}\mathit{t}}$ in 6, verify the differentiation formula 7.

Using the mass-spring analogy, predict the behavior as $\mathbf{t}\to \infty$ of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem.

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{100}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem $\left(11\right)$ in Example $\mathbf{3}$ by taking $\mathbf{b}\mathbf{=}\mathbf{0}$.

  a)   If $\mathbf{m}\mathbf{=}\mathbf{10}\mathbf{ }\mathbf{kg}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{250}\mathbf{ }\mathbf{kg}\mathbf{/}{\mathbf{sec}}^2\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{m}$, and $\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{sec}$, find the     equation of motion for this undamped vibrating spring.

b)  After how many seconds will the mass in part $\left(a\right)$  first cross the equilibrium point?

c)   When the equation of motion is of the form displayed in $\left(9\right)$, the motion is said to be oscillatory with frequency $\mathbf{\beta }\mathbf{/}\mathbf{2}\mathbf{\pi }$. Find the frequency of oscillation for the spring system of part $\left(a\right)$.

Vibrating Spring with Damping. Using the model for a vibrating spring with damping discussed in Example$\mathbf{3}$

$\left(a\right)$ Find the equation of motion for the vibrating spring with damping if  $\mathbf{m}\mathbf{=}\mathbf{10}\mathbf{ }\mathbf{kg}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{60}\mathbf{ }\mathbf{kg}\mathbf{/}\mathbf{sec}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{250}\mathbf{ }\mathbf{kg}\mathbf{/}{\mathbf{sec}}^2\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{m}\mathbf{,}$and  $\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{sec}$.

$\left(b\right)$ After how many seconds will the mass in part $\left(a\right)$ first cross the equilibrium point?

$\left(c\right)$ Find the frequency of oscillation for the spring system of part $\left(a\right)$. 

 $\left(d\right)$Compare the results of problems $\mathbf{32}$ and  $\mathbf{33}$ determine what effect the damping has on the frequency of oscillation. What other effects does it have on the solution?

$\mathbf{RLC}$Series Circuit. In the study of an electrical circuit consisting of a resistor, capacitor, inductor, and an electromotive force (see Figure), we are led to an initial value problem of the form

$\begin{array}{r} \mathbf{\left(}\mathbf{20}\mathbf{\right)}\mathbf{L}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{+}\mathbf{RI}\mathbf{+}\frac{q}{C}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{;}\\ \mathbf{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{q}}_0\\ \mathbf{I}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{I}}_0\mathbf{,}\end{array}$

where $\mathbf{L}$ is the inductance in henrys, $\mathbf{R}$ is the resistance in ohms, $\mathbf{C}$ is the capacitance in farads,  $\mathbf{E}\left(t\right)$ is the electromotive force in volts,$\mathbf{q}\left(t\right)$  is the charge in coulombs on the capacitor at the time $\mathbf{t}$, and   $\mathbf{I}\mathbf{=}\mathbf{dq}\mathbf{/}\mathbf{dt}$ is the current in amperes. Find the current at time t if the charge on the capacitor is initially zero, the initial current is zero,  $\mathbf{L}\mathbf{=}\mathbf{10}\mathbf{H}\mathbf{,}\mathbf{R}\mathbf{=}\mathbf{20}\mathbf{\Omega }\mathbf{,}\mathbf{C}\mathbf{=}\mathbf{(}\mathbf{6260}{\mathbf{)}}^{-1}\mathbf{ }\mathbf{F}$ , and  $\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{ }\mathbf{V}$ . 

Swinging Door. The motion of a swinging door with an adjustment screw that controls the amount of friction on the hinges is governed by the initial value problem

$\mathbf{I\theta }\text{'}\text{'}\mathbf{+}\mathbf{b\theta }\text{'}\mathbf{+}\mathbf{k\theta }\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_0\mathbf{,}\mathbf{\theta }\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{v}}_0$ ,

where $\mathbf{\theta }$ is the angle that the door is open, $\mathbf{I}$ is the moment of inertia of the door about its hinges, $\mathbf{b}\mathbf{>}\mathbf{0}$ is a damping constant that varies with the amount of friction on the door, $\mathbf{k}\mathbf{>}\mathbf{0}$ is the spring constant associated with the swinging door,${\mathbf{\theta }}_{\mathbf{0}}$  is the initial angle that the door is opened, and ${\mathbf{v}}_{\mathbf{0}}$ is the initial angular velocity imparted to the door (see figure). If  $\mathrm{I}$ and $\mathbf{k}$  are fixed, determine for which values of $\mathbf{b}$  the door will not continually swing back and forth when closing.

Although the real general solution form (9) is convenient, it is also possible to use the form (21) ${\mathit{d}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }\mathbf{)}\mathbf{t}}\mathbf{+}{\mathit{d}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{(}\mathbf{\alpha }\mathbf{-}\mathbf{i\beta }\mathbf{)}\mathbf{t}}$ to solve initial value problems, as illustrated in Example 1. The coefficients ${\mathit{d}}_{\mathbf{1}} \mathbf{\text{and}} {\mathit{d}}_{\mathbf{2}}$ are complex constants.

1. Use the form (21) to solve Problem 21. Verify that your form is equivalent to the one derived using (9).
2. Show that, in general, ${\mathit{d}}_{\mathbf{1}} \mathbf{\text{and}} {\mathit{d}}_{\mathbf{2}}$ in (21) must be complex conjugates so that the solution is real.

The auxiliary equations for the following differential equations have repeated complex roots. Adapt the "repeated root" procedure of Section $\mathbf{4}\mathbf{.}\mathbf{2}$  to find their general solutions:

$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{y}\text{'}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

$\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{y}\text{'}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{4}\mathbf{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{12}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{16}\mathbf{y}\text{'}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{0}$

Prove the sum of angles formula for the sine function by following these steps. Fix $\mathbf{x}$ .

$\left(a\right)$  Let  $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{)}$. Show that $\mathbf{f}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$, the standard sum of angles formula for $\mathbf{sin}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{)}$ . $\mathbf{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{sinx}$ , and $\mathbf{f}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{cosx}$.

 $\left(b\right)$ Use the auxiliary equation technique to solve the initial value problem $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{sinx}$, and $\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{cosx}$ 

$\left(c\right)$ By uniqueness, the solution in part $\left(b\right)$  is the same as following these steps. Fix $\mathbf{x}$. $\mathbf{f}\left(\mathbf{t}\right)$ from part $\left(a\right)$. Write this equality; this should be the standard sum of angles formula for sin$\left(x+t\right)$.

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution to the given equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{t}}^{-1}{\mathbf{e}}^t$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution to the given equation.$\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{t}}^3\mathbf{cos}\mathbf{4}\mathbf{t}$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathrm{sinx}}{{\mathbf{e}}^{4x}}$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation. $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{=}{\mathbf{3}}^t$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{-}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{sec\theta }$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.$\mathbf{2}\mathbf{\omega }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{\omega }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{4}{\mathbf{xsin}}^2\mathbf{x}\mathbf{+}\mathbf{4}{\mathbf{xcos}}^2\mathbf{x}$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.$\mathbf{8}\mathbf{z}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{2}\mathbf{z}\left(\mathbf{x}\right)\mathbf{=}\mathbf{3}{\mathbf{x}}^{100}{\mathbf{e}}^{4x}\mathbf{cos}\mathbf{25}\mathbf{x}$

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation. $\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{sin}\mathbf{3}\mathbf{t}$