Q9E

Question

Let $ϕ (x)$ denote the solution to the initial value problem

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

⦁ Show that $ϕ (x) = 1 - ϕ' (x) = 1 - x + ϕ (x)$

⦁ Argue that the graph of $ϕ$ is decreasing for x near zero and that as x increases from zero, $ϕ (x)$ 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 $y = ϕ (x)$ crosses the line $y = x$ .Consider the sign of $ϕ (x *)$ and argue that $ϕ$ has a relative minimum at x*.

⦁ What can you say about the graph of $y = ϕ (x)$ for x > x*?

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

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

⦁ Sketch the solution $y = ϕ (x)$ using the direction field in part (f).

Step-by-Step Solution

Verified

Answer

⦁ Proved

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

⦁ The graph of $y = ϕ (x)$ increases.

⦁ The graph of $ϕ (x)$ increases for x > x*

⦁ Proved

⦁ The graph is drawn below

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

Given, $ϕ (x)$ is a solution to the problem then

$ϕ' (x) = x - ϕ (x)$

Differentiate concerning x,

$ϕ'' (x) = 1 - ϕ' (x) = 1 - (x - ϕ (x)) = 1 - x + ϕ (x),$

Hence it is shown that $ϕ (x) = 1 - ϕ' (x) = 1 - x + ϕ (x)$

22(b): Analyze values of ϕ ' ( x )

Given that $ϕ (x)$ is a solution to the initial value problem,

$ϕ' (x) = x - ϕ (x), ϕ (0) = 1 {\frac{d ϕ}{dx}|}_{x = 0} = 0 - ϕ (0) = 0 - 1 = - 1 < 0$

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 $ϕ (x)$ keeps on decreasing until $ϕ' (x)$ is negative,

$x - ϕ (x) < 0 x < ϕ (x)$

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

Hence, $ϕ (x)$ decreases until it crosses the line y = x.

33(c): Compute ϕ ' ' ( x * )

From (b), it is clear that the point where $ϕ (x)$ crosses the line y = x is a critical point.

From (a), $ϕ'' (x *) = 1 - ϕ' (x *) = 1 - 0 = 1 > 0$

Hence, By Second Derivative Test, it can be concluded that $ϕ (x)$ has a relative minimum at x*.

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

From (c), x* is the point where $ϕ (x *) = x *$ , i.e., $ϕ (x)$ has a relative minimum at x*.

For x > x*, the graph of $ϕ (x)$ increases. (By Monotonicity Test)

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

Consider, $y = x - 1$

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

And,

$x - y = x - (x - 1) = x - x + 1 = 1$

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

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