Question

Let $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$denote the solution to the initial value problem

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

⦁ Show that $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{+}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$

⦁ Argue that the graph of $\varphi$ is decreasing for x near zero and that as x increases from zero, $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$decreases until it crosses the line y = x, where its derivative is zero.

⦁ Let x* be the abscissa of the point where the solution curve $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ crosses the line $y=x$.Consider the sign of $\varphi \mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)}$ and argue that $\varphi$ has a relative minimum at x*.

⦁ What can you say about the graph of $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for x > x*?

⦁ Verify that y = x – 1 is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ and explain why the graph of $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ always stays above the line $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$.

⦁ Sketch the direction field for $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ by using the method of isoclines or a computer software package.

⦁ Sketch the solution $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ using the direction field in part (f).

Short answer

⦁ Proved

⦁ $\varphi$is decreasing near x = 0 until it crosses the line y = x.

⦁ The graph of $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$increases.

⦁ The graph of $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$increases for x > x*

⦁ Proved

⦁ The graph is drawn below

⦁ The graph is drawn below

Step-by-step solution

1(a): Show that  ϕ(x)=1-ϕ'(x)=1-x+ϕ(x)

Given, $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is a solution to the problem then

$\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{-}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Differentiate concerning x,

$$\begin{gathered} \varphi \mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{=}\mathbf{1}\mathbf{-}\mathbf{(}\mathbf{x}\mathbf{-}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{+}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,} \end{gathered}$$

Hence it is shown that $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{+}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$

2(b): Analyze values of  ϕ'(x)

Given that $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$is a solution to the initial value problem,

$$\begin{gathered} \varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{-}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\varphi \mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \\ {\left.\frac{\mathbf{d}\varphi }{\mathbf{dx}}\right|}_{\mathbf{x}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{-}\varphi \mathbf{\left(}\mathbf{0}\mathbf{\right)} \\ \mathbf{=}\mathbf{0}\mathbf{-}\mathbf{1} \\ \mathbf{=}\mathbf{-}\mathbf{1}\mathbf{<}\mathbf{0} \end{gathered}$$

Now, as the derivative is negative at x = 0, it can be concluded that the function itself is decreasing near x = 0.

Next, the function $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$keeps on decreasing until $\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is negative,

$$\begin{gathered} \mathbf{x}\mathbf{-}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{<}\mathbf{0} \\ \mathbf{x}\mathbf{<}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

By Intermediate Value Theorem, this $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$keeps on decreasing until it reaches the point where $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$, (i.e., $x=y$).

Hence, $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$decreases until it crosses the line y = x.

3(c): Compute  ϕ''(x*)

From (b), it is clear that the point where $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$crosses the line y = x is a critical point.

From (a), $$\begin{gathered} \varphi \mathbf{\text{'}}\mathbf{\text{'}}\mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{-}\varphi \mathbf{\text{'}}\mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{-}\mathbf{0} \\ \mathbf{=}\mathbf{1}\mathbf{>}\mathbf{0} \end{gathered}$$

Hence, By Second Derivative Test, it can be concluded that role="math" localid="1663925734266" $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ has a relative minimum at x*.

4(d): Analyze the graph of  ϕ(x) for x > x*

From (c), x* is the point where $\mathit{\varphi }\mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{*}$, i.e., $\mathit{\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ has a relative minimum at x*.

For x > x*, the graph of $\mathit{\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$increases. (By Monotonicity Test)

5(e): Verify y = x-1 is a solution to  dydx=x-y and apply the existence and uniqueness of the solution theorem

Consider, $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$

Then, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{1}$

And,

$$\begin{gathered} \mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)} \\ \mathbf{=}\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{1} \end{gathered}$$

Thus, y = x - 1 is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$

As $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$and $\frac{\partial \mathbf{f}}{\mathbf{dy}}\mathbf{=}\mathbf{-}\mathbf{1}$are continuous on the whole plane. The initial value problem, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{(}{\mathbf{x}}_{\mathbf{0}}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{0}}$, has a unique solution for any ${\mathbf{x}}_{\mathbf{0}}$and ${\mathbf{y}}_{\mathbf{0}}$(Using Existence and Uniqueness of Solutions Theorem).

Therefore, the curve $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$always stays above the line $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$

6(f): Sketch the direction field

7(g): Sketch the solution y=ϕ(x) .

By putting the values of x in equation I get the graph.