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Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 4

September 24, 2014 etoix Leave a comment

Problem

Determine whether the equations on the right define implicit functions of x. For those which do, determine whether they are implicit solutions of the differential equations on the left.

$y^2 - 1 - (2y + xy) y\prime = 0$	$y^2 - 1 = (x+2)^2$
$e^{x-y} + e^{y-x} \frac{dy}{dx} = 0$	$e^{2y}+e^{2x} = 1$
$\frac{dy}{dx} = -\frac{y}{x}$	$x^2 + y^2 + 1 = 0$

Solution

We start with $y^2 - 1 = (x+2)^2$ .

Solve for y and get $y = \pm \sqrt{(x+2)^2 + 1}$ .

So we know that this function is defined for all $\mathbb{R}$ .

Let implicitly differentiate and get:

$2yy\prime = 2(x+2)$

Simplfy:

$yy\prime = x+2$

The derivative is $y\prime = \frac{x+2}{y}$ .

Let’s substitute into $y^2 - 1 - (2y + xy) y\prime = 0$ to see if we get identity.

$y^2 - 1 - (2y + xy ) \frac{x+2}{y} = 0$

$y^2 - 1 - (2[x+2] + x[x+2]) = 0$

$y^2 - 1 - (x^2 + 4x + 4) = 0$

Now let’s substitute with $y^2 - 1 = (x+2)^2$ .

$latex (x+2)^2 – (x^2 + 4x + 4) = 0$

We have identity, so $y^2 - 1 = (x+2)^2$ does implicitly solve $y^2 - 1 - (2y + xy) y\prime = 0$ .

Keep in mind that y has two solutions: $y = \pm \sqrt{(x+2)^2 + 1}$ . So you need to pick one to complete the answer.

We start with $e^{2y}+e^{2x} = 1$ .

$e^{anything} > 0$ . In order for two positive real numbers to add up to 1, they both need to be less than 1.

Below is a plot:

Notice that the plot approaches both axes asymptotically.

The domain is $x < 0$ . Here is where I disagree with the answer key, which assert $x \neq 0$ .

Let’s implicitly differentiate and get:

$2 e^{2y} y\prime + 2 e^{2x} = 0$

Simplify:

$e^{2y} y\prime + e^{2x} = 0$

Divide by $e^y$ .

$e^{y} y\prime + e^{2x-y} = 0$

Divide by $e^x$ .

$e^{y-x} y\prime + e^{x-y} = 0$

This is exactly the differential equation we’re looking for, so $e^{2y}+e^{2x} = 1$ is a solution for $e^{x-y} + e^{y-x} \frac{dy}{dx} = 0$ .

Let’s start with $x^2 + y^2 + 1 = 0$ .

This is $x^2 + y^2 = -1$ .

$x^2$ and $y^2$ both must be positive and cannot be -1.

Therefore, this implicit function is not defined in $\mathbb{R}$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 3

August 29, 2014 etoix Leave a comment

Problem

Show that the differential equation

$|\frac{dy}{dx}| + |y| + 1 = 0$

has no solutions.

Solution

The range of $|\frac{dy}{dx}| \geq 0$ .

The range of $|y| \geq 0$ .

So the minimum of the left side is $0 + 0 + 1$ .

Hence, we have no identity.

Therefore, a solution is impossible.

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 2

August 28, 2014 etoix Leave a comment

Problem

Prove that the function sin the right-hand column below are solutions of the differential equation sin the left-hand columns [sic]. (Be sure to state the common interval for which solution and differential equation make sense.)

$y^{\prime} + y = 0$	$y = e^{-x}$
$y^{\prime} = e^x$	$y = e^x$
$\frac{d^2y}{dx^2} = \frac{1}{\sqrt{1-x^2}}$	$y = x \arcsin{x} + \sqrt{1 - x^2}$
$f^{\prime}(x) = f^{\prime\prime}(x)$	$y = e^x + 2$
$xy^{\prime} = 2y$	$y = x^2$
$(1 + x^2) y^{\prime} = xy$	$y = \sqrt{1 + x^2}$
$\cos(\theta) \frac{dr}{d\theta} - 2r \sin{\theta} = 0$	$r = a \sec^2(\theta)$
$y^{\prime\prime} - y = 0$	$y = a e^x + b e^{-x}$
$f^{\prime}(x) = \frac{1}{3} f(x)$	$f(x) = 2 e^{\frac{x}{3}}$
$xy^{\prime} + y = y^2$	$y = \frac{2}{x+2}$
$x + y y^{\prime} = 0$	$y = \sqrt{16 - x^2}$

Solution

$y = e^{-x}$

$y^{\prime} = -e^{-x}$ .

$-e^{-x} + e^{-x} = 0$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = e^x$

$y^{\prime} = e^x$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = x \arcsin{x} + \sqrt{1 - x^2}$

$\frac{dy}{dx} = \arcsin(x) + \frac{x}{\sqrt{1-x^2}} - \frac{2x}{2 \sqrt{1-x^2}}$

$\frac{dy}{dx} = \arcsin(x)$

$\frac{d^2y}{dx^2} = \frac{1}{\sqrt{1-x^2}}$

The differential equation and solution are valid for $-1 \leq x \leq 1$ .

$y = e^x + 2$

$y^{\prime} = e^x$

$y^{\prime\prime} = e^x$

$f^{\prime}(x) = f^{\prime\prime}(x)$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = x^2$

$y^{\prime} = 2x$

$x (2x) = 2 ( x^2)$

The differential equation and solution are valid for $-\infty < x < \infty$ .

The answer key says that the domain is $x \neq 0$ . It is unclear why this is so.

$y = \sqrt{1 + x^2}$

$y^{\prime} = \frac{2x}{2 \sqrt{1+x^2}}$

$(1 + x^2) y^{\prime} = xy$

$(1 + x^2) ( \frac{2x}{2 \sqrt{1+x^2}} ) = x ( \sqrt{1 + x^2} )$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$r = a \sec^2(\theta)$

$\frac{dr}{d\theta} = 2 a \sec^2(\theta) \tan(\theta)$

$\cos(\theta) \frac{dr}{d\theta} - 2r \sin{\theta} = 0$

$\cos(\theta) ( 2 a \sec^2(\theta) \tan(\theta) ) - 2 ( a \sec^2(\theta)) \sin(\theta) = 0$

$2 a \sec^2(\theta) \sin(\theta) - 2 ( a \sec^2(\theta)) \sin(\theta) = 0$

The differential equation is valid for $-\infty < \theta < \infty$ but the solution is valid for $\theta \neq \frac{\pi}{2}, -\frac{\pi}{2}, \frac{3 \pi}{2}, - \frac{3 \pi}{2}, \frac{5 \pi}{2}, - \frac{5 \pi}{2}, \dots$ .

$y = a e^x + b e^{-x}$

$y^{\prime} = a e^x - b e^{-x}$

$y^{\prime\prime} = a e^x + b^{-x}$

$a e^x + b^{-x} - ( a e^x + b^{-x} ) = 0$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$f(x) = 2 e^{\frac{x}{3}}$

$f^{\prime}(x) = \frac{2 e^{\frac{x}{3}}}{3}$

$f^{\prime}(x) = \frac{1}{3} f(x)$

$\frac{2 e^{\frac{x}{3}}}{3} = \frac{1}{3} (2 e^{\frac{x}{3}} )$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = \frac{2}{x+2}$

$y^{\prime} = - \frac{2}{(x+2)^2}$

$xy^{\prime} + y = y^2$

$x ( - \frac{2}{(x+2)^2} ) + \frac{2}{x+2} = ( \frac{2}{x+2} ) ^2$

$- \frac{2x}{ (x+2)^2 } + \frac{2x+4}{ (x+2)^2} = \frac{4}{ ( x+ 2)^2}$

The differential equation is valid for $-\infty < x < \infty$ but the solution is valid for $x \neq -2$ .

The answer keys says that the domain is $x \neq -2, 0$ . It is unclear why 0 is a problem. The solution is defined for $x = 0$ .

$y = \sqrt{16 - x^2}$

$y^{prime} = - \frac{2x}{2 \sqrt{16 - x^2}}$

$x + y y^{\prime} = 0$

$x - \sqrt{16 - x^2} ( \frac{2x}{2 \sqrt{16 - x^2}} ) = 0$

$x - x = 0$

The differential equation is valid for $-\infty < x < \infty$ but the solution is valid for $-4 < x < 4$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 1

August 27, 2014 etoix Leave a comment

Problem

Determine the order of the following differential equations.

$dy + ( xy - \cos{x} ) dx = 0$
$y^{\prime\prime} + x y^{\prime\prime} + 2y ( y^{\prime} )^3 + xy = 0$
$(\frac{d^2y}{dx^2})^3 - (y^{\prime\prime\prime})^4 + x = 0$
$e^{y^{\prime\prime\prime}} + x y^{\prime\prime} + y = 0$

Solution

$dy + ( xy - \cos{x} ) dx = 0$

This is order 1.

$y^{\prime\prime} + x y^{\prime\prime} + 2y ( y^{\prime} )^3 + xy = 0$

This is order 2.

$(\frac{d^2y}{dx^2})^3 - (y^{\prime\prime\prime})^4 + x = 0$

This is order 3.

$e^{y^{\prime\prime\prime}} + x y^{\prime\prime} + y = 0$

This is order 3.

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 17

August 26, 2014 etoix Leave a comment

Problem

Define a function of three independent variables $x_1, x_2, x_3$ ; of n independent variables $x_1 \dots, x_n$ . Hint: See Definition 2.6.

Solution

Definition 2.6:

If to each element (x,y) of a set E in the plane (the set must be specified) there corresponds one and only one real value of z, then z is said to be a function of x and y for the set E. In this event, x,y are called independent variables and z a dependent variable.

A function of three independent variables:

$f(x_1, x_2, x_3) = x_1 + x_2 + x_3$

A function of n independent variables:

$f(x_1, \dots, x_n) = x_1 + \dots + x_n$

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 16

August 25, 2014 etoix 1 Comment

Problem

Can you apply the method of implicit differentiation as taught in the calculus to the function of problem 14, of problem 15?

Solution

The answer key says no, but this is clearly possible.

$\frac{d}{dx} ( \sqrt{x^2- y^2} + \arccos{\frac{x}{y}} ) = \frac{2x - 2y \frac{dy}{dx}}{2 \sqrt{x^2 - y^2}} - \frac{- \frac{x \frac{dy}{dx}}{y^2} + \frac{1}{y}}{ \sqrt{1 - \frac{x^2}{y^2}}}$

$\frac{d}{dx} ( \sqrt{x^2- y^2} + \arccos{\frac{x}{y}} ) = \frac{2x - 2y \frac{dy}{dx}}{2 \sqrt{x^2 - y^2}} + \frac{- \frac{x \frac{dy}{dx}}{y^2} + \frac{1}{y}}{ \sqrt{1 - \frac{x^2}{y^2}}}$

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 15

August 23, 2014 etoix Leave a comment

Problem

Explain why

$\sqrt{x^2 - y^2} + \arcsin{\frac{x}{y}} = 0$

does not define y as an implicit function of x.

Solution

The domain of $\arcsin{z}$ is $-1 \leq z \leq 1$ .

So $-1 \leq \frac{x}{y} \leq 1$ . Hence, $-y \leq x \leq y$ .

Now consider $\sqrt{x^2 - y^2}$ . To stay within $\mathbb{R}$ , it must be true that $x^2 \geq y^2$ .

For the most part $-y \leq x \leq y$ contradicts $x^2 \geq y^2$ . The only case where both of these constraints holds true is when $y = x$ . Therefore, $y = g(x) = x$ is the only candidate to satisfy this equation.

$\sqrt{x^2 - y^2} + \arcsin{\frac{x}{y}} = 0$

$\sqrt{x^2 - x^2} + \arcsin{\frac{x}{x}} = 0$

$\sqrt{0} + \arcsin{1} = 0$

$0 + \frac{\pi}{2} = 0$

We have reached a contradiction, therefore the equation does not define y as an implicit function of x.

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 14

August 22, 2014 etoix 2 Comments

Problem

Find the function g(x) that is implicitly defined by the relation

$\sqrt{x^2 - y^2} + \arccos{\frac{x}{y}} = 0$ , $y \neq 0$ .

Solution

The domain of $\arccos{z}$ is $-1 \leq z \leq 1$ .

So $-1 \leq \frac{x}{y} \leq 1$ . Hence, $-y \leq x \leq y$ .

Now consider $\sqrt{x^2 - y^2}$ . To stay within $\mathbb{R}$ , it must be true that $x^2 \geq y^2$ .

$\sqrt{x^2 - y^2} + \arccos{\frac{x}{y}} = 0$

$\sqrt{x^2 - x^2} + \arccos{\frac{x}{x}} = 0$

$\sqrt{0} + \arccos{1} = 0$

$0 + 0 = 0$

We claim that $g(x) = x$ allows y to be implicitly defined by x.

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 13

August 21, 2014 etoix Leave a comment

Problem

Explain why the procedure followed in problem 12, applied to the relation $x^2 + y^2 + 1 = 0$ and yielding the result

$\frac{dy}{dx} = - \frac{x}{y}$

is meaningless.

Solution

The answer key simply says that “ $x^2 + y^2 + 1 = 0$ does not define a function.” While this is true, I feel that their solution is incomplete. After all, we’ve seen in problem 12 that a non-function equation can be turn into a function simply by restricting the domain or sectioning off parts of the graph.

With $x^2 + y^2 + 1 = 0$ however, we have something different. This particular equation’s domain has no elements in $\mathbb{R}$ at all. There is no domain to restriction. There is no graph to section off. It is simply impossible in $\mathbb{R}$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 12

August 20, 2014 etoix Leave a comment

Problem

The following is a standard type of exercise in the calculus.

If $x^3 y^3 - 3xy = 0$ , then $3x^2 + 3y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} - 3y = 0$ . Therefore

$\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}$ , $y^2 \neq x$

Explain by the use of Fig. 2.77 what this means geometrically.

Solution

As discussed in problem 11, the Folium of Descartes is not a function but an equation. We can graphically section off parts of the plot to transform the equation into a function.

The manipulation above is simply the derivative so geometrically $\frac{y - x^2}{y^2 - x}$ is the slope of the equation.