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Show that any function of the form $$x=C_{1} \cosh \omega t+C_{2} \sinh \omega t$$ satisfies the differential equation $$x^{\prime \prime}-\omega^{2} x=0. $$

Short Answer

Expert verified
The function satisfies the differential equation after differentiating and substituting.

Step by step solution

01

Differentiation of the Function

To solve this exercise, we first need to differentiate the function \( x(t) = C_1 \cosh(\omega t) + C_2 \sinh(\omega t) \) once with respect to \( t \). The derivative of the hyperbolic cosine function is hyperbolic sine and vice versa, thus:\[ x'(t) = C_1 \omega \sinh(\omega t) + C_2 \omega \cosh(\omega t). \]
02

Second Differentiation

Next, differentiate \( x'(t) \) to find the second derivative \( x''(t) \). Applying the derivatives of hyperbolic functions again gives:\[ x''(t) = C_1 \omega^2 \cosh(\omega t) + C_2 \omega^2 \sinh(\omega t). \]
03

Equation Substitution

Substitute \( x(t) \) and \( x''(t) \) into the differential equation \( x''(t) - \omega^2 x(t) = 0 \). This results in:\[ C_1 \omega^2 \cosh(\omega t) + C_2 \omega^2 \sinh(\omega t) - \omega^2 \left(C_1 \cosh(\omega t) + C_2 \sinh(\omega t)\right) = 0. \]
04

Simplification

Simplify the expression by distributing the \( \omega^2 \) term and combining like terms:\[ \omega^2 C_1 \cosh(\omega t) + \omega^2 C_2 \sinh(\omega t) - \omega^2 C_1 \cosh(\omega t) - \omega^2 C_2 \sinh(\omega t) = 0. \]Both terms cancel each other out, simplifying to:\[ 0 = 0. \]
05

Verification

Since we obtained a true statement \( 0 = 0 \), this verifies that the original function \( x(t) = C_1 \cosh(\omega t) + C_2 \sinh(\omega t) \) does indeed satisfy the differential equation \( x'' - \omega^2 x = 0 \).

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but for hyperbolas rather than circles. They are widely used in mathematics, especially in solving certain types of differential equations. The most common hyperbolic functions are hyperbolic sine (\( \sinh \)) and hyperbolic cosine (\( \cosh \)).

These functions can be defined using exponential functions, which makes them particularly useful in calculus and differential equations:
  • Hyperbolic sine: \( \sinh x = \frac{e^x - e^{-x}}{2} \)
  • Hyperbolic cosine: \( \cosh x = \frac{e^x + e^{-x}}{2} \)
Interestingly, hyperbolic functions satisfy similar properties to trigonometric functions. For example, \( \cosh^2 x - \sinh^2 x = 1 \), paralleling the identity \( \cos^2 x + \sin^2 x = 1 \).

In differential equations, hyperbolic functions often arise naturally as solutions, particularly for equations that resemble oscillatory motion—except with exponential growth or decay characteristics.
Second Derivative
The second derivative of a function provides us with the rate at which the rate of change of a function is changing. In simpler terms, it's like looking at the acceleration in physics—the speed at which velocity changes. For our specific function \( x(t) = C_1 \cosh(\omega t) + C_2 \sinh(\omega t) \), finding the second derivative involves differentiating twice.

The first differentiation gives us the velocity-like term: \( x'(t) = C_1 \omega \sinh(\omega t) + C_2 \omega \cosh(\omega t) \).

Differentiating again, for the acceleration (the second derivative), we get: \( x''(t) = C_1 \omega^2 \cosh(\omega t) + C_2 \omega^2 \sinh(\omega t) \).

This step is crucial because it helps verify if a given function is the solution to a differential equation by checking its relationship with the original function.
Solution Verification
Verifying a solution means checking if the proposed function indeed satisfies a given differential equation. It's like proving the proposed answer is correct. For our problem, we want to see if \( x(t) = C_1 \cosh(\omega t) + C_2 \sinh(\omega t) \) satisfies \( x''(t) - \omega^2 x(t) = 0 \).

We substitute \( x(t) \) and its second derivative into the differential equation. What we observe is that each term in the differential equation cancels out when doing the substitution and simplification:
  • The second derivative term: \( C_1 \omega^2 \cosh(\omega t) + C_2 \omega^2 \sinh(\omega t) \)
  • Subtraction of \( \omega^2 x(t) \): \( \omega^2 \times (C_1 \cosh(\omega t) + C_2 \sinh(\omega t)) \)
After simply aligning terms on both sides, each component negates the other, resulting in \( 0 = 0 \). This confirms that we have a true solution, which means our function correctly solves the differential equation.
Linear Homogeneous Differential Equations
A linear homogeneous differential equation is one where all terms are proportional to the function and its derivatives, and there are no external (non-zero) terms—no constant terms or forcing functions. The equation \( x'' - \omega^2 x = 0 \) is a classic example.

Such equations are important because their solutions generally form a vector space. In simpler terms, any non-zero solution can be multiplied by a constant to produce another solution, and any two solutions can be added together to form a new solution.

Our function \( x(t) = C_1 \cosh(\omega t) + C_2 \sinh(\omega t) \) is a linear combination of hyperbolic functions. It represents the general solution to the given linear homogeneous differential equation, where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.

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