Question

Consider the differential equation \(y^{\prime \prime}(t)-k^{2} y(t)=0,\) where \(k>0\) is a real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} e^{t}+C_{2} e^{-t} .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} e^{2 t}+C_{2} e^{-2 t}\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture. d. For a positive real number \(k\), verify that the general solution of the equation may also be expressed in the form \(y(t)=C_{1} \cosh k t+C_{2} \sinh k t,\) where cosh and sinh are the hyperbolic cosine and hyperbolic sine, respectively (Section \(7.3)\)

Short answer

Question: Verify that the suggested solutions are true for the given differential equation \(y^{\prime\prime}(t) - k^2 y(t) = 0\) and provide the general solutions for arbitrary k. a. Show that \(y(t) = C_1e^t + C_2e^{-t}\) is a solution when \(k=1\). b. Show that \(y(t) = C_1e^{2t} + C_2e^{-2t}\) is a solution when \(k=2\). c. Find the general solution for arbitrary \(k > 0\). d. Express the general solution using hyperbolic sine and cosine functions. Answer: a. The given solution \(y(t) = C_1e^t + C_2e^{-t}\) is verified as the correct solution when \(k=1\). b. The given solution \(y(t) = C_1e^{2t} + C_2e^{-2t}\) is verified as the correct solution when \(k=2\). c. The general solution for arbitrary \(k > 0\) is \(y(t) = C_1e^{kt} + C_2e^{-kt}\). d. The general solution can also be expressed as \(y(t) = C_1\cosh(kt) + C_2\sinh(kt)\).

Step-by-step solution

a. Verifying the solution when k=1

Firstly, we are given the differential equation \(y^{\prime\prime}(t) - k^2 y(t) = 0\) and the suggested solution \(y(t)=C_1e^t + C_2e^{-t}\). To verify by substitution, we will take the first and second derivatives of the proposed solution, and then plug them back into the differential equation. First derivative: \(y^{\prime}(t) = C_1e^t - C_2e^{-t}\) Second derivative: \(y^{\prime\prime}(t) = C_1e^t + C_2e^{-t}\) Now, plug the function and its second derivative back into the differential equation with \(k=1\): \((C_1e^t + C_2e^{-t}) - (C_1e^t + C_2e^{-t}) = 0\) Simplifying, we get: \(0 = 0\) So, the given solution \(y(t) = C_1e^t + C_2e^{-t}\) is verified as the correct solution when \(k=1\).

b. Verifying the solution when k=2

Now, we are given the differential equation \(y^{\prime\prime}(t) - k^2 y(t) = 0\) and the suggested solution \(y(t)=C_1e^{2t} + C_2e^{-2t}\). We will use the same method of substitution as before. First derivative: \(y^{\prime}(t) = 2C_1e^{2t} - 2C_2e^{-2t}\) Second derivative: \(y^{\prime\prime}(t) = 4C_1e^{2t} + 4C_2e^{-2t}\) Now, plug the function and its second derivative back into the differential equation with \(k=2\): \((4C_1e^{2t} + 4C_2e^{-2t}) - 4(C_1e^{2t} + C_2e^{-2t}) = 0\) Simplifying, we get: \(0 = 0\) So, the given solution \(y(t) = C_1e^{2t} + C_2e^{-2t}\) is verified as the correct solution when \(k=2\).

c. General solution for arbitrary k

Let the general solution be \(y(t) = C_1e^{kt} + C_2e^{-kt}\). To verify this for arbitrary \(k > 0\), we will take its first and second derivatives and substitute them into the given differential equation. First derivative: \(y^{\prime}(t) = kC_1e^{kt} - kC_2e^{-kt}\) Second derivative: \(y^{\prime\prime}(t) = k^2C_1e^{kt} + k^2C_2e^{-kt}\) Now, plug the function and its second derivative back into the differential equation: \((k^2C_1e^{kt} + k^2C_2e^{-kt}) - k^2(C_1e^{kt} + C_2e^{-kt}) = 0\) Simplifying, we get: \(0 = 0\) So, the general solution for the given differential equation for arbitrary \(k > 0\) is \(y(t) = C_1e^{kt} + C_2e^{-kt}\).

d. Expressing the general solution using hyperbolic functions

Given the general solution \(y(t) = C_1e^{kt} + C_2e^{-kt}\), we will show that it can also be expressed as \(y(t) = C_1\cosh(kt) + C_2\sinh(kt)\), where \(\cosh\) and \(\sinh\) are hyperbolic cosine and sine functions, respectively. Recall that \(\cosh(x) = \frac{e^x + e^{-x}}{2}\) and \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). Multiplying these definitions by \(C_1\) and \(C_2\), and summing them, we get: \(C_1\cosh(kt) + C_2\sinh(kt) = \frac{C_1(e^{kt}+e^{-kt})}{2} + \frac{C_2(e^{kt}-e^{-kt})}{2}\) Combining terms, we get: \(y(t) = C_1e^{kt} + C_2e^{-kt}\) This shows that the general solution can also be expressed as \(y(t) = C_1\cosh(kt) + C_2\sinh(kt)\).

Key concepts

General Solution

The term 'general solution' in the context of differential equations is critically important, as it encapsulates all possible solutions to a differential equation. It incorporates arbitrary constants because these equations represent families of curves, rather than a single curve, on a graph. For a second-order linear differential equation like \(y^{{\prime \prime}}(t) - k^2 y(t) = 0\), where \((k > 0)\), the general solution involves two arbitrary constants, often denoted as \((C_1)\) and \((C_2)\). \
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\The general solution is representative of the system's behavior for all initial conditions, capturing various potential scenarios that the system could encounter. For instance, the general solution for \((k = 1)\), as given in the exercise, is \((y(t) = C_1 e^t + C_2 e^{-t}\)), which means that for any specific initial conditions, you could find the corresponding values of \((C_1)\) and \((C_2)\) to fit those conditions. \
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\When dealing with general solutions for differential equations, verification requires substituting the proposed solution back into the original equation. If the equation is satisfied, as in the steps provided, that confirms the general solution is correct. \

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• Different values of \((k)\) affect the shape and nature of the solution but don't change the fundamental approach to finding the general solution.
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• For any \((k > 0)\), the structure of the solution remains consistent with the exponents of 'e' being \((\pm kt)\).
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• This approach is generic and can be applied to various second-order linear differential equations with constant coefficients.

Hyperbolic Functions

Hyperbolic functions, such as the hyperbolic cosine \(\cosh\) and hyperbolic sine \(\sinh\), often appear in solutions for differential equations and have remarkable properties similar to their trigonometric counterparts. These functions relate to the exponential function, and they can be expressed as: \

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• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
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• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)

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\These functions describe the shape of a hanging cable or chain, known as a catenary, and turn up in various physical systems. In our exercise, we see a relationship between the general solution of the differential equation for arbitrary \((k > 0)\) and these hyperbolic functions. \
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\The expressions for these functions make them particularly useful when consolidating the general solution that includes terms like \((e^{kt}\)) and \((e^{-kt}\)). Therefore, you can represent the general solution by combining \(\cosh\) and \(\sinh\), which makes the solution not only succinct but also highlights the natural appearance of these functions in physical phenomena and mathematical theory. By using hyperbolic functions, we can often simplify complex calculations and provide insights into the behavior of the solutions.

Second-order Linear Differential Equation

A second-order linear differential equation with constant coefficients is a fundamental type of equation in mathematics, particularly in physics and engineering. Such an equation takes the form \(y^{{\prime \prime}} + py'+ qy = 0\) or sometimes without the first derivative, as \(y^{{\prime \prime}} - k^2 y = 0\) like in our exercise. Here, \((p\)) and \((q\)) are constants, and \((k\)) is a real positive number representing the coefficient in the equation. \
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\The solutions to these equations involve exponential functions or, as seen in part (d) of the exercise, hyperbolic functions. They often model scenarios where the current state of the system is influenced by its immediate past behavior, which is typical in mechanical systems like oscillations and electric circuits. \
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\One crucial aspect of solving these equations is understanding their corresponding characteristic equation, which must be solved to find the general solution. The characteristic equation for the given differential equation would be \((r^2 - k^2 = 0\)), leading to solutions involving \((e^{kt}\)) and \((e^{-kt}\)). \

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• The second-order nature indicates that the solution's behavior is dependent on both its initial position and velocity (or analogous quantities).
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• When solved, the general solution includes two arbitrary constants, reflecting that the equation's second-order nature requires two initial conditions to be specified.
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• The linearity ensures the principle of superposition applies, which means that the sum of any two solutions is itself a solution.

\Identifying and solving such differential equations is a key skill in applied mathematics, offering insights into a wide range of dynamic systems.