Question

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

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

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

Figure 4.3.7 Graph for problem 48

Short answer

The differential equation in point$\left(b\right)y\text{'}\text{'}+4y=0$belongs to the given graph because this wave is a wave with a big frequency with a small period$\mathrm{\pi }$.

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 big frequency with a small period $\mathrm{\pi }$, then we can say that the differential equation in $\left(b\right)y\text{'}\text{'}+4y=0$ 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

$m^2+4=0$

Then we have the roots

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

Then we can have the solution as

$$\begin{gathered} y=k_1{e}^{m_{1}x}+k_2{e}^{m_{2}x} \\ y=k_1{e}^{2ix}+k_2{e}^{-2ix} \\ y=k_1(\cos 2x+\sin 2x)+k_2(\cos 2x-\sin 2x) \\ y=(k_1+k_2)\cos 2x+(k_1-k_2)\sin 2x \\ y=c_1\cos 2x+c_2\sin 2x \end{gathered}$$

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 $\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$belongs to the given graph because this wave is a wave with a big frequency with a small period$\mathbf{\pi }$.