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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: Q14 E (page 14)

Show that $蠒 (x) = c_{1} \sin x + c_{2} \cos x,$ is a solution to $\frac{d^{2} y}{{dx}^{2}} + y = 0$ for any choice of the constants $c_{1}$ and $c_{2}$ . Thus, $c_{1} \sin x + c_{2} \cos x,$ is a two-parameter family of solutions to the differential equation.

Short Answer

Expert verified

$蠒 (x) = c_{1} \sin x + c_{2} \cos x,$ is a solution to $\frac{d^{2} y}{{dx}^{2}} + y = 0$ , for any choice of the constants $c_{1}$ and $c_{2}$ .

$c_{1} \sin x + c_{2} \cos x,$ is a two-parameter family of solutions to the differential equation.

Step by step solution

01

Taking the given function as y

First of all, take the given function as,

$蠒 (x) = y$

02

Differentiating the given function concerning x

Differentiating concerning x,

$\frac{dy}{dx} = c_{1} \cos x - c_{2} \sin x$

Again, differentiating concerning x,

$\frac{d^{2} y}{{dx}^{2}} = - c_{1} \sin x - c_{2} cosx$

03

Simplification of the differential equation obtained in step 2

In step 2, we get $\frac{d^{2} y}{{dx}^{2}} = - c_{1} sinx - c_{2} cosx$

$\begin{array}{l} \frac{d^{2} y}{{dx}^{2}} = - (c_{1} \sin x + c_{2} \cos x) \\ \frac{d^{2} y}{{dx}^{2}} = - y \\ \frac{d^{2} y}{{dx}^{2}} + y = 0 \end{array}$

Which is identical to the given differential equation.

Hence $蠒 (x) = c_{1} \sin x + c_{2} \cos x,$ is a solution to $\frac{d^{2} y}{{dx}^{2}} + y = 0$ , for any choice of the constants $c_{1}$ and $c_{2}$ . Therefore, $c_{1} \sin x + c_{2} \cos x$ is a two-parameter family of solutions to the given differential equation.

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Most popular questions from this chapter

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\frac{dy}{dx} = 3 x - \sqrt[3]{y - 1}, y (2) = 1$

In Project C of Chapter 4, it was shown that the simple pendulum equation $胃'' (t) + 蝉颈苍胃 (t) = 0$ has periodic solutions when the initial displacement and velocity show that the period of the solution may depend on the initial conditions by using the vectorized Runge鈥揔utta algorithm with h= 0.02 to approximate the solutions to the simple pendulum problem on

[0, 4] for the initial conditions:

localid="1664100454791" $(a) 鈥� 鈥� 胃 (0) = 0 . 1, 胃' (0) = 0 (b) 鈥� 鈥� 胃 (0) = 0 . 5, 胃' (0) = 0 (c) 鈥� 鈥� 胃 (0) = 1 . 0, 胃' (0) = 0$

[Hint: Approximate the length of time it takes to reach].

In Problem 19, solve the given initial value problem

$\begin{array}{l} y''' 鈭� y'' 鈭� 4 y' + 4 y = 0 \\ y (0) = 鈭� 4 \\ y' (0) = 鈭� 1 \\ y'' (0) = 鈭� 19 \end{array}$

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

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

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

Spring Pendulum.Let a mass be attached to one end of a spring with spring constant kand the other end attached to the ceiling. Let $l_{o}$ be the natural length of the spring, and let l(t) be its length at time t. If $胃 (t)$ is the angle between the pendulum and the vertical, then the motion of the spring pendulum is governed by the system

$l'' (t) - l (t) 胃' (t) - 驳肠辞蝉胃 (t) + \frac{k}{m} (l - l_{o}) = 0 l^{2} (t) 胃'' (t) + 2 l (t) l' (t) 胃' (t) + gl (t) 蝉颈苍胃 (t) = 0$

Assume g = 1, k = m = 1, and $l_{o}$ = 4. When the system is at rest, $l = l_{o} + \frac{mg}{k} = 5$ .

a. Describe the motion of the pendulum when $l (0) = 5 . 5, l' (0) = 0, 胃 (0) = 0, 胃' (0) = 0$ .

b. When the pendulum is both stretched and given an angular displacement, the motion of the pendulum is more complicated. Using the Runge鈥揔utta algorithm for systems with h = 0.1 to approximate the solution, sketch the graphs of the length l and the angular displacement u on the interval [0,10] if $l (0) = 5 . 5, l' (0) = 0, 胃 (0) = 0 . 5, 胃' (0) = 0$ .

See all solutions

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