Question

The graph of a solution of a second-order initial-value problem ${\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}{\mathit{y}}_{\mathbf{0}}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}{\mathit{y}}_{\mathbf{1}}$, is given in Figure 1.R.1. Use the graph to estimate the values of ${\mathit{y}}_{\mathbf{0}}$and ${\mathit{y}}_{\mathbf{1}}$.

Short answer

$y_0=-3\mathrm{and}{y}_1=0$

Step-by-step solution

Existence of a Unique solution.

Let R be a rectangular region in the xy-plane defined by$a\le x\le b,c\le y\le d$ that contains the point $(x_0,y_0)$in its interior. If$f(x,y)$ and $\partial f/\partial y$are continuous on R, then there exists some interval $I_0:(x_0-h,x_0+h),h>0$ contained in$[a,b]$ and a unique function y(x), defined on $I_0$, that is a solution of the initial-value problem (2).

Using the given graph

From the given solution graph, we see that$y\left(2\right)=-3$.

So,

$\begin{array}{c}y\left(2\right)=y_0\\ y_0=-3\end{array}$

Now,$y\text{'}\left(2\right)$ will be the slope of the tangent line to the given solution graph at $x=2$, which we clearly see is a horizontal line and so will have a slope equal to zero. Hence,

$\begin{array}{c}y\text{'}\left(2\right)=0\\ y_1=0\end{array}$

Therefore, the values are $y_0=-3\mathrm{and}{y}_1=0$.