Problem 32 - Essential Calculus

Question

Show that the function satisfies the differential equation. \(y=a \cosh x\) \(y^{\prime \prime}-y=0\)

Step-by-Step Solution

Answer

The function \(y=a \cosh x\) satisfies the differential equation \(y^{\prime \prime}-y=0\).

Step 1: Calculating the first derivative of the function

Use the chain rule to calculate the derivative of \(y = a \cosh x\). The derivative of \(\cosh x\) is \(\sinh x\), so the derivative of the function is \(y' = a \sinh x\).

Step 2: Calculating the second derivative of the function

The derivative of \(\sinh x\) is \(\cosh x\). So, the second derivative of our original function is \(y'' = a \cosh x\).

Step 3: Substituting into the Differential Equation

Having calculated the second derivative of the function, substitute the initial function and its second derivative into the given differential equation \(y^{\prime \prime}-y=0\). If the equation holds true with these values, then the function satisfies the differential equation.

Step 4: Verification

Upon substituting, we get \(y^{\prime \prime}-y = a \cosh x - a \cosh x = 0\). Thus, the differential equation holds and we have demonstrated that the function \(y=a \cosh x\) satisfies the differential equation.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, such as \( ext{cosh}(x)\) and \( ext{sinh}(x)\), are analogs to trigonometric functions but are based on hyperbolas rather than circles. They are particularly useful in solving certain types of differential equations and modeling phenomena that display hyperbolic patterns.\

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• Cosh \((x)\): The function \( ext{cosh}(x) = \frac{e^x + e^{-x}}{2}\) behaves similarly to \( ext{cos}(x)\), showing symmetry about the y-axis and having a minimum value at \(x = 0\).
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• Sinh \((x)\): The function \( ext{sinh}(x) = \frac{e^x - e^{-x}}{2}\) is akin to \( ext{sin}(x)\), being an odd function and having zeros at \(x = 0\).
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\Hyperbolic functions satisfy many similarities and identities akin to trigonometric functions. For example, \( ext{cosh}^2(x) - ext{sinh}^2(x) = 1\) mirrors the circular identity \( ext{cos}^2( heta) + ext{sin}^2( heta) = 1\). These functions become quite useful in various areas of mathematics including differential equations.

Second Order Differential Equations

A second-order differential equation involves the second derivative of an unknown function. In the context of this exercise, the equation is \(y^{\prime\prime} - y = 0\). This type is classified as linear and homogeneous.\

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• Linear: The equation is in a linear form because the unknown function and its derivatives appear in a linear manner with constant coefficients.
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• Homogeneous: It means all terms involve the function or its derivatives, and the right-hand side is zero. These equations usually have solutions that can be expressed as linear combinations of functions.
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\The given second order differential equation is verified through substitution once the derivatives are calculated correctly. Confirming the function \(y = a \cosh x\) as a solution displays the role of hyperbolic functions as it reappears in its own second derivative form. This behavior is a characteristic property crucial in verifying the solution to the differential equation.

Chain Rule

The chain rule is a fundamental tool in calculus for finding derivatives of composite functions. In this exercise, it is used to differentiate the function \(y = a \cosh x\). Understanding how the chain rule is applied can simplify complex differentiation processes.\\For a composite function \(f(g(x))\), the chain rule states that the derivative is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function: \(f'(g(x)) \cdot g'(x)\).\\In our exercise: \

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• The function \(y = a \cosh x\) requires finding a derivative where the outer function is \(a\cdot \text{cosh}(x)\).
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• The inner function, \(g(x) = x\), implies that \(g'(x) = 1\), simplifying the process.
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\The chain rule simplifies to taking the derivative of \(\cosh x\), giving \(\sinh x\). By following these steps carefully, one ensures the correct derivatives are used in solving differential equations.