ODE by Tenenbaum & Pollard, Exercise 3, Problem 2

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 $\theta \neq \frac{\pi}{2}, -\frac{\pi}{2}, \frac{3 \pi}{2}, - \frac{3 \pi}{2}, \frac{5 \pi}{2}, - \frac{5 \pi}{2}, \dots$ .