Q9E Let 蠒(x)聽denote the solution t... [FREE SOLUTION]

91影视

Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 1: Q9E (page 1)

Let $蠒 (x)$ denote the solution to the initial value problem
$\frac{dy}{dx} = x - y, y (0) = 1$
猞� Show that $蠒 (x) = 1 - 蠒' (x) = 1 - x + 蠒 (x)$
猞� Argue that the graph of $蠒$ is decreasing for x near zero and that as x increases from zero, $蠒 (x)$ decreases until it crosses the line y = x, where its derivative is zero.
猞� Let x* be the abscissa of the point where the solution curve $y = 蠒 (x)$ crosses the line $y = x$ .Consider the sign of $蠒 (x )$ and argue that $蠒$ has a relative minimum at x.
猞� What can you say about the graph of $y = 蠒 (x)$ for x > x*?
猞� Verify that y = x 鈥� 1 is a solution to $\frac{dy}{dx} = x - y$ and explain why the graph of $y = 蠒 (x)$ always stays above the line $y = x - 1$ .
猞� Sketch the direction field for $\frac{dy}{dx} = x - y$ by using the method of isoclines or a computer software package.
猞� Sketch the solution $y = 蠒 (x)$ using the direction field in part (f).

Short Answer

Expert verified

猞� Proved

猞� $蠒$ is decreasing near x = 0 until it crosses the line y = x.

猞� The graph of $y = 蠒 (x)$ increases.

猞� The graph of $蠒 (x)$ increases for x > x*

猞� Proved

猞� The graph is drawn below

Step by step solution

1(a): Show that ϕ(x)=1-ϕ'(x)=1-x+ϕ(x)

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

$蠒' (x) = x - 蠒 (x)$

Differentiate concerning x,

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

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

2(b): Analyze values of ϕ'(x)

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

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

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

$x - 蠒 (x) < 0 x < 蠒 (x)$

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

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

3(c): Compute ϕ''(x*)

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

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

Hence, By Second Derivative Test, it can be concluded that role="math" localid="1663925734266" $蠒 (x)$ has a relative minimum at x*.

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

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

For x > x*, the graph of $蠒 (x)$ 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, $y = x - 1$

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 $\frac{鈭� f}{dy} = - 1$ 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).