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Fundamentals Of Differential Equations And Boundary Value Problems

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

$Math Studyset Vaia Explanations$ Math

9th Edition · 616 pages

ISBN: 9780321977069

Chapter 1: Q19 E (page 14)

Show that the equation ${(\frac{dy}{dx})}^{2} + y^{2} + 4 = 0$ has no (real-valued) solution.

Short Answer

Expert verified

${(\frac{dy}{dx})}^{2} + y^{2} + 4 = 0$ has no (real-valued) solution.

Step by step solution

01

Simplification of the given differential equation

$\begin{array}{l} {(\frac{d y}{d x})}^{2} = - y^{2} - 4 \\ {(\frac{d y}{d x})}^{2} = - (y^{2} + 4) \\ \frac{d y}{d x} = \sqrt{- (y^{2} + 4)} \end{array}$

02

Determining if the given equation has a real-valued solution or not

Now from Step 1, this is clear that the value of $\frac{d y}{d x}$ is not real.

Thus ${(\frac{dy}{dx})}^{2} + y^{2} + 4 = 0$ has no (real-valued) solution.

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Most popular questions from this chapter

The logistic equation for the population (in thousands) of a certain species is given by $\frac{dp}{dt} = 3 p - 2 p^{2}$ .

⦁ Sketch the direction field by using either a computer software package or the method of isoclines.

⦁ If the initial population is 3000 [that is, p(0) = 3], what can you say about the limiting population?

⦁ If $p (0) = 0 . 8$ , what is $\lim_{t \to + \infty} p (t)$ ?

⦁ Can a population of 2000 ever decline to 800?

Decide whether the statement made is True or False. The function $y (x) = - \frac{1}{3} (x + 1)$ is a solution to $\frac{dy}{dx} = \frac{y - 1}{x + 3}$ .

Determine which values of m the function $ϕ (x) = e^{mx}$ is a solution to the given equation.

(a) $\frac{d^{2} y}{{dx}^{2}} + 6 \frac{dy}{dx} + 5 y = 0$

(b) $\frac{d^{3} y}{{dx}^{3}} + 3 \frac{d^{2} y}{{dx}^{2}} + 2 \frac{dy}{dx} = 0$

Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{s + 1}{{(s^{2} + 1)}^{2}}$

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$x = 2 \cos t - 3 \sin t, x'' + x = 0$

See all solutions

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