/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 In each part, verify that the fu... [FREE SOLUTION] | 91Ó°ÊÓ

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In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. \(y^{\prime \prime}+4 y=0\) (a) \(\sin 2 x\) and \(\cos 2 x\) (b) \(c_{1} \sin 2 x+c_{2} \cos 2 x\left(c_{1}, c_{2} \text { constants }\right)\)

Short Answer

Expert verified
Functions in both parts satisfy the differential equation after substitution.

Step by step solution

01

Differentiate the Functions in Part (a)

First, we need to find the second derivative of the function \(y = \sin 2x\). The first derivative is \(y' = 2\cos 2x\). The second derivative is then \(y'' = -4\sin 2x\).Next, for \(y = \cos 2x\), the first derivative is \(y' = -2\sin 2x\) and the second derivative is \(y'' = -4\cos 2x\).
02

Substitute into the Differential Equation (Part a)

Now, substitute \(y = \sin 2x\) and its second derivative into the equation.Substitute in the differential equation: \(y'' + 4y = -4\sin 2x + 4\sin 2x = 0\).Do the same for \(y = \cos 2x\), substitute: \(y'' + 4y = -4\cos 2x + 4\cos 2x = 0\).Both functions satisfy the differential equation.
03

Differentiate the General Combination Function (Part b)

Consider the function \(y = c_1 \sin 2x + c_2 \cos 2x\) where \(c_1\) and \(c_2\) are constants.Find the first derivative: \(y' = 2c_1 \cos 2x - 2c_2 \sin 2x\).Find the second derivative: \(y'' = -4c_1 \sin 2x - 4c_2 \cos 2x\).
04

Substitute the General Function into the Differential Equation (Part b)

Substitute the expression for \(y\) and \(y''\) into the differential equation:\(y'' + 4y = (-4c_1 \sin 2x - 4c_2 \cos 2x) + 4(c_1 \sin 2x + c_2 \cos 2x)\).This simplifies to \(-4c_1 \sin 2x - 4c_2 \cos 2x + 4c_1 \sin 2x + 4c_2 \cos 2x = 0\).Thus, for any values of \(c_1\) and \(c_2\), the differential equation is satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
The second derivative is core to solving many differential equations. It helps in determining how a function curves, or changes its rate of change.
Imagine you're driving; the first derivative is your speed, while the second derivative is how your speed is changing—an acceleration or deceleration.
##### Deriving the Second DerivativeIn calculus, the second derivative of a function refers to the derivative of the derivative.
  • For a function like \(y = \sin 2x\), the first derivative \(y'\) is the rate at which \(y\) changes with respect to \(x\), so \(y' = 2\cos 2x\).
  • By differentiating \(y'\) once more, we derive the second derivative \(y'' = -4\sin 2x\).
The process is similar for other functions, and calculating these accurately allows us to solve the differential equation by substitution. Understanding this can ultimately help predict the behavior of physical systems, finance models, or population dynamics.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, play a significant role in solving differential equations, especially those involving harmonic motion.
These functions are periodic, meaning they repeat their values in regular intervals, making them ideal for modeling cycles such as waves, or oscillations.##### Sine and Cosine Functions
  • For \(y = \sin 2x\), the function deals with oscillations that repeat every \(\pi\) radians since its period is reduced from the default \(2\pi\) by a factor of \(2\).
  • Similarly, \(y = \cos 2x\) shares these characteristics, albeit with a phase shift.
The derivatives of these functions transform sine into cosine and vice versa, with additional constants derived from the chain rule.
This symmetry and transformation trait make trigonometric functions especially useful in solving linear differential equations, as they often bring about simplifications.
General Solutions
In differential equations, finding a general solution means determining an expression that satisfies the equation for any given constants.
This provides a family of solutions rather than a single solution, offering more flexibility.
##### General Solution FrameworkConsider the differential equation: - \(y^{\prime \prime} + 4y = 0\).- Solutions can be taken as combinations of sine and cosine functions, adjusted by constants, such as \(y = c_1 \sin 2x + c_2 \cos 2x\).
  • The constants \(c_1\) and \(c_2\) can be determined given initial conditions or specific requirements.
  • By substituting back into the differential equation, \((-4c_1 \sin 2x - 4c_2 \cos 2x + 4(c_1 \sin 2x + c_2 \cos 2x)) = 0\), we confirm that this form holds true.
Understanding general solutions expands our ability to customize solutions, making them applicable to different scenarios or models without redefining the entire function.

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Most popular questions from this chapter

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