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 $\mathbb{R}$ .

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