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

Short answer

The solution (a) $\mathbf{y}\left(\mathbf{1}\right)=\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{1}\right)=-\mathbf{2}$matches the given curve.

Step-by-step solution

Step 1: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 see that the graph passes through $(1,1)$, which corresponds to the following condition.

$y\left(1\right)=1$

From graph ,$\mathbf{y}\text{'}\left(\mathbf{1}\right)$is negative.

It seems that the condition matches either (a) or (c) since$y\left(1\right)=1$. We check the slope of the function at $x=1$, which is negative.

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

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

We check the slope given in (a) and (c). We see that only (a) gives information of negative slope.

Therefore, the solution (a)$\mathbf{y}\left(\mathbf{1}\right)=\mathbf{1},\mathbf{y}\text{'}\left(\mathbf{1}\right)=-\mathbf{2}$ matches the given curve.