Question

Question: In problems43–48 each figure represents the graph of a particularsolution of one of the following differential equations:

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

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

Figure 4.3.4 Graph for problem

Short answer

The differential equation in point $\left(e\right) y+2y\text{'}+2y=0$ belongs to the given graph because the value of y tends to 0 when the value of x increases (i.e., there is an exponential function with the trigonometric function).

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 the value of y tends to 0 when the value of x increases (i.e., there is an exponential function with the trigonometric function), then we can say that the differential equation in $\left(e\right) y+2y\text{'}+2y=0$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

$m^2+2m+2=0$

Then we have the roots

$m_{1,2}=\frac{-2\pm \sqrt{4-4\times 2}}{2}=-1\pm i$

Then we can have the solution as

$\begin{array}{rcl}y& =& k_1{e}^{m_{1}x}+k_2{e}^{m_{2}x}\\ & =& k_1{e}^{\left(-1+i\right)x}+k_2{e}^{\left(-1-i\right)x}\\ & =& k_1{e}^{-x}\left(\cos x+i\sin x\right)+k_2{e}^{-x}\left(\cos x-i\sin x\right)\\ & =& \left(k_1+k_2\right)e^{-x}\cos x+\left(k_1-k_2\right)i{e}^{-x}\sin x\\ y& =& c_1{e}^{-x}\cos x+c_2{e}^{-x}\sin x\end{array}$

Where, $c_1=k_1+k_2$and $c_2=\left(k_1-k_2\right)i$are arbitrary constants.

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

Therefore, the differential equation in point $\left(e\right) y+2y\text{'}+2y=0$belongs to the given graph because the value ofytends to0when the value ofxincreases (i.e., there is an exponential function with the trigonometric function).