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 that 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.

Short answer

#Answer# a. Verified that when \(k=1\), a solution of the equation is \(y(t)=C_1 e^t + C_2 e^{-t}\). b. Verified that when \(k=2\), the general solution of the equation is \(y(t)=C_1 e^{2t} + C_2 e^{-2t}\). c. The general solution for the equation with arbitrary \(k>0\) is \(y(t) = C_1 e^{kt} + C_2 e^{-kt}\). d. The general solution can also be expressed in terms of hyperbolic functions as \(y(t) = C'_1 \cosh kt + C'_2 \sinh kt\).

Step-by-step solution

Substitute the given solution into the differential equation

We are given the solution as \(y(t) = C_1 e^t + C_2 e^{-t}\). We need to compute the second derivative and substitute it back into the differential equation. The first and second derivatives are \(y^{\prime}(t) = C_1 e^t - C_2 e^{-t}\), and \(y^{\prime\prime}(t) = C_1 e^t + C_2 e^{-t}\), respectively. Now, we substitute these derivatives and the original function into the differential equation, \(y^{\prime\prime}(t) - k^2y(t)=0\).

Check if the equation is satisfied

With the given solution \(y(t)=C_1 e^t + C_2 e^{-t}\) and given that \(k=1\), we have: $$ y^{\prime\prime}(t) - k^2y(t) = (C_1 e^t + C_2 e^{-t}) - (C_1 e^t + C_2 e^{-t}) = 0 $$ This clearly satisfies the equation, as the left side becomes zero. Therefore, we have verified that when \(k=1\), a solution of the equation is \(y(t)=C_1 e^t + C_2 e^{-t}\). #b. Verifying the solution for k=2#

Substitute the given solution into the differential equation

We are given the solution as \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\). We need to compute the second derivative and substitute it back into the differential equation. The first and second derivatives are \(y^{\prime}(t) = 2C_1 e^{2t} - 2C_2 e^{-2t}\), and \(y^{\prime\prime}(t) = 4C_1 e^{2t} + 4C_2 e^{-2t}\), respectively. Now, we substitute these derivatives and the original function into the differential equation, \(y^{\prime\prime}(t) - k^2y(t)=0\).

Check if the equation is satisfied

With the given solution \(y(t)=C_1 e^{2t} + C_2 e^{-2t}\) and given that \(k=2\), we have: $$ y^{\prime\prime}(t) - k^2y(t) = (4C_1 e^{2t} + 4C_2 e^{-2t}) - 4(C_1 e^{2t} + C_2 e^{-2t}) = 0 $$ This clearly satisfies the equation, as the left side becomes zero. Therefore, we have verified that when \(k=2\), the general solution of the equation is \(y(t)=C_1 e^{2t} + C_2 e^{-2t}\). #c. Finding the general solution for arbitrary \(k>0\)#

Propose a general solution

Based on parts a and b, we can propose a general solution for any \(k>0\) as follows: \(y(t) = C_1 e^{kt} + C_2 e^{-kt}\).

Substitute the general solution into the differential equation

We compute the first and second derivatives of the proposed general solution and get: \(y^{\prime}(t) = kC_1 e^{kt} - kC_2 e^{-kt}\), and \(y^{\prime\prime}(t) = k^2C_1 e^{kt} + k^2C_2 e^{-kt}\). Now, we substitute these derivatives and the original function into the differential equation: $$ y^{\prime\prime}(t) - k^2y(t) = (k^2C_1 e^{kt} + k^2C_2 e^{-kt}) - k^2(C_1 e^{kt} + C_2 e^{-kt}) = 0 $$ The equation is satisfied for arbitrary \(k>0\), so the general solution for the equation is indeed \(y(t) = C_1 e^{kt} + C_2 e^{-kt}\). #d. Expressing the general solution in terms of hyperbolic functions#

Define hyperbolic sine and cosine

The hyperbolic sine and cosine functions are defined as follows: $$ \sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}. $$

Rewrite the general solution using hyperbolic functions

We can rewrite the general solution as: $$ y(t) = \frac{C_1 + C_2}{2} e^{kt} + \frac{C_1 - C_2}{2} e^{-kt} = \frac{C_1 + C_2}{2} \cosh kt + \frac{C_1 - C_2}{2} \sinh kt. $$ We can rename the constants, letting \(C'_1 = \frac{C_1 + C_2}{2}\) and \(C'_2 = \frac{C_1 - C_2}{2}\), giving the general solution in the desired form: $$ y(t) = C'_1 \cosh kt + C'_2 \sinh kt. $$ Thus, we have verified that the general solution of the equation may also be expressed in the form \(y(t) = C_1 \cosh kt + C_2 \sinh kt\).

Key concepts

General Solution

In differential equations, a general solution refers to a solution that encompasses all possible solutions of the equation given any initial conditions or parameters. For second-order linear homogeneous differential equations like the one provided, the general solution typically involves two arbitrary constants. These constants account for any specific characteristics needed if initial conditions are provided later.

In the exercise's differential equation, we began by assuming the solution was of the form \( y(t) = C_1 e^{kt} + C_2 e^{-kt} \). This form is derived from the natural exponential solutions to linear differential equations, driven by the fact that the derivative of an exponential function is proportionate to the function itself. Therefore, when you solve the characteristic equation associated with the differential equation, you end up with solutions that include \( e^{kt} \) and \( e^{-kt} \).

The constants \( C_1 \) and \( C_2 \) are placeholders for any specific initial conditions that might further define the solution. In the absence of these, the general solution remains as a combination of these individual solutions.

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are important in the context of differential equations and often appear when these are expressed in exponential forms. The hyperbolic sine and cosine functions are defined as:

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

By understanding these definitions, it becomes clear how hyperbolic functions relate to the exponential functions used in solving differential equations.

In our specific problem, the general solution of the differential equation could be expressed using hyperbolic functions by combining the exponential terms. The conversion from \( y(t) = C_1 e^{kt} + C_2 e^{-kt} \) to \( y(t) = C_1 \cosh kt + C_2 \sinh kt \) is achieved by manipulating and re-expressing exponential terms using the definitions above.

These transformations allow one to exploit properties of hyperbolic functions, such as simplicity in certain integrations or visualizing solutions to physical systems that exhibit hyperbolic behavior, such as catenary curves or certain relativistic systems.

Substitution Method

The substitution method in solving differential equations is a technique used to verify potential solutions by substituting them back into the original equation. This process involves checking if the proposed solution satisfies the differential equation under the given conditions.

For example, in initial steps, you find the derivatives of the proposed solution, perform necessary calculations, and substitute these back. For our exercise, this meant substituting \( y(t) = C_1 e^{kt} + C_2 e^{-kt} \) and its derivatives into the original equation, \( y''(t) - k^2y(t) = 0 \).

By calculating \( y''(t) \) and replacing \( y(t) \), we check if the left side of the equation indeed results in zero, confirming that \( y(t) \) is a solution for the differential equation.

This technique is crucial because it ensures the solution's validity without ambiguity, providing certainty that the proposed form addresses the differential behavior encoded by the original equation. By systematically verifying parts, one can gain an intuitive understanding of how the solution structures relate to differential properties expressed in mathematical form.