Question

In Problems 35–38 the graph of a member of a family of solutions of a second-order differential equation ${\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{)}$is given. Match the solution curve with at least one pair of the following initial conditions.

(a) $\mathbf{y}\left(\mathbf{1}\right)=\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{1}\right)=-\mathbf{2}$

(b) $\mathbf{y}(-\mathbf{1})=\mathbf{0},\mathbf{y}\text{'}(-\mathbf{1})=-\mathbf{4}$

(c) $\mathbf{y}\left(\mathbf{1}\right)=\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{1}\right)=\mathbf{2}$

(d) $\mathbf{y}\left(\mathbf{0}\right)=-\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{0}\right)=\mathbf{2}$

(e) $\mathbf{y}\left(\mathbf{0}\right)=-\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{0}\right)=\mathbf{0}$

(f) $\mathbf{y}\left(\mathbf{0}\right)=-\mathbf{4},\mathbf{y}\text{'}\left(\mathbf{0}\right)=-\mathbf{2}$

FIGURE 1.2.7 Graph for Problem 35

Short answer

The solutions (b) $\mathbf{y}(-\mathbf{1})=\mathbf{0},\mathbf{y}\text{'}(-\mathbf{1})=-\mathbf{4}$ and (f) $\mathbf{y}\left(\mathbf{0}\right)=-\mathbf{4},\mathbf{y}\text{'}\left(\mathbf{0}\right)=-\mathbf{2}$ match the given curve.

Step-by-step solution

Definition of differential equation

Differential Equation is defined as the equation that consists of the derivatives of one or more dependent functions in respect to one or more independent functions.

Check the slope of the function

We can see that the graph passes through (-1,0) and (0,-4) , which corresponds to the following conditions.

$$\begin{gathered} y(-1)=0 \\ y\left(0\right)=-4 \end{gathered}$$

It seems that the condition matches both (b) and (f) since $y-(-1)=0$ and $y\left(0\right)=-4$. We check the slope of the function at $x=-1$ and $x=0$, which are negative.

From the graph, both $y\text{'}(-1)$ and $y\text{'}\left(0\right)$ are negative.

We check the slopes given in (b) and (f) and they make sense because they are negative. Therefore, we choose both (b) and (f).

$y\text{'}(-1)=-4$ from (b)$y\text{'}\left(0\right)=-2$ and from (f).

Therefore, the solutions (b) $\mathbf{y}(-\mathbf{1})=\mathbf{0},\mathbf{y}\text{'}(-\mathbf{1})=-\mathbf{4}$ and (f) $y\left(0\right)=-4,y\text{'}\left(0\right)=-2$ match the given curve.