Question

In problems 43–48 each figure represents the graph of a particularsolution of one of the following differential equations:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

Match a solution curve with one of the differential equations. Explain your reasoning.

Figure 4.3.6 Graph for problem 47

Short answer

The differential equation in point$\left(d\right)y\text{'}\text{'}+y=0$ belongs to the given graph because this wave is a wave with a small frequency with a big period$2\mathrm{\pi }$.

Step-by-step solution

Find the solution curve

• First, the graph shown in the figure for this problem in the book is a graph for a trigonometric function (i.e., complex roots), then we can say that the differential equations in points, (a) (c) and (f) do not belong here, since they contain real roots.
• Second, this graph shows that this wave is a wave with a small frequency with a big period $2\mathrm{\pi }$, then we can say that the differential equation in $\left(d\right)y\text{'}\text{'}+y=0$belongs to this graph.

Solve the differential equation

Third, we can solve this differential equation as the following:

From the D.E, we can obtain the auxiliary equation as

$m^2+1=0$

Then we have the roots

$m_{1,2}=\pm i$

Then we can have the solution as

Where $c_1=k_1+k_2$and $c_2=k_1-k_2$are an arbitrary constant.

Finally, we can substitute with any value for in this solution to be sure that we are right.

Therefore, the differential equation in point $\mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$belongs to the given graph because this wave is a wave with a small frequency with a big period$\mathbf{2}\mathbf{\pi }$