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 4.3.2 Graph for problem 43

Short answer

The differential equation in point $\left(f\right) y-3y\text{'}+2y=0$ belongs to the given graph because the value of y tends to 0 when the value of x is negative.

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 an exponential function (i.e., real roots), then we can say that the differential equations in points, (b), (d), and (e) do not belong here, since they contain complex roots.

• Second, this graph shows that the value of y is positive whatever the value of x is positive or negative, then we can exclude the differential equation in point (c) since it has a repeated root and contains y as a negative.
• Third, this graph shows that the value of y tends to 0 when the value of x is negative, then we can say that the differential equation in $\left(f\right) y-3y\text{'}+2y=0$belongs to this graph.

Solve the differential equation

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

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

$m^2-3m+2=0$

Then,$(m-1)(m-2)=0$

Then we have the roots

$m_1=1 \text{and} m_2=2$

Then we can have the solution as

role="math" localid="1667996603027" $$\begin{gathered} y=k_1{e}^{m_{1}x}+k_2{e}^{m_{2}x} \\ =k_1{e}^x+k_2{e}^{2x} \end{gathered}$$

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 $\left(f\right) y-3y\text{'}+2y=0$belongs to the given graph because the value ofytends to0when the value ofxis negative.