Question

v\(4 y^{\prime \prime}-y=0 ; \quad y_{1}(t)=e^{-t / 2}, \quad y_{2}(t)=-2 e^{t / 2} ; \quad \bar{y}(t)=2 \cosh (t / 2)\)

Short answer

#Question# Show that the functions \(y_1(t) = e^{-t/2}\), \(y_2(t) = -2e^{t/2}\), and \(\bar{y}(t) = 2\cosh(t/2)\) are solutions to the given second order differential equation \(4y'' - y = 0\). Moreover, demonstrate that \(\bar{y}(t)\) is a linear combination of \(y_1(t)\) and \(y_2(t)\).

Step-by-step solution

Verify the solutions y_1(t) and y_2(t)

We are given two functions \(y_1(t) = e^{-t/2}\) and \(y_2(t) = -2e^{t/2}\). Let's verify if they are solutions to the given ODE: \(4y'' - y = 0\) Firstly, compute the first and second order derivatives for each function: For \(y_1(t) = e^{-t/2}\), \(y_1'(t) = \frac{-1}{2} e^{-t/2}\) \(y_1''(t) = \frac{1}{4} e^{-t/2}\) For \(y_2(t) = -2e^{t/2}\), \(y_2'(t) = -e^{t/2}\) \(y_2''(t) = \frac{1}{2} e^{t/2}\) Now let's substitute each function and its derivatives into the ODE and check if the equation holds true: For \(y_1(t)\): \(4( \frac{1}{4} e^{-t/2}) - e^{-t/2} = e^{-t/2} - e^{-t/2} = 0\) Similarly for \(y_2(t)\): \(4( \frac{1}{2} e^{t/2}) + 2e^{t/2} = 2e^{t/2} + 2e^{t/2} = 4e^{t/2} - 4e^{t/2} = 0\) Since both functions satisfy the given ODE, they are valid solutions.

Demonstrate that \(\bar{y}(t)\) is a linear combination of solutions and verify it as a solution to the ODE

Now let's consider the function \(\bar{y}(t) = 2\cosh(t/2)\). We want to express this function as a linear combination of the two solutions \(y_1(t)\) and \(y_2(t)\). We know that \(\cosh(t/2) = \frac{e^{t/2} + e^{-t/2}}{2}\), therefore \(\bar{y}(t) = e^{t/2} + e^{-t/2}\) Notice that \(\bar{y}(t)\) can be expressed as a linear combination of \(y_1(t)\) and \(y_2(t)\): \(\bar{y}(t) = \frac{1}{2} y_2(t) - \frac{1}{2} y_1(t)\) Now let's verify that \(\bar{y}(t)\) is a solution to the given ODE. Compute the derivatives: \(\bar{y}'(t) = \frac{1}{2} e^{t/2} - \frac{1}{2} e^{-t/2}\) \(\bar{y}''(t) = \frac{1}{4} e^{t/2} + \frac{1}{4} e^{-t/2}\) Substitute the function and derivatives in the ODE: \(4( \frac{1}{4} e^{t/2} + \frac{1}{4} e^{-t/2}) - (e^{t/2} + e^{-t/2}) = 0\) Simplifying, \( e^{t/2} + e^{-t/2} - (e^{t/2} + e^{-t/2}) = 0\) Thus, \(\bar{y}(t)\) is a solution to the given ODE. In conclusion, we have verified that functions \(y_1(t)\) and \(y_2(t)\) are solutions to the given ODE, and that function \(\bar{y}(t)\) is a linear combination of these two solutions and also a solution to the given ODE.

Key concepts

Verification of solutions

Verification of solutions is an important step when dealing with ordinary differential equations (ODEs). It involves substituting candidate solutions back into the original equation to determine whether they satisfy it. For the ODE in question, which is \(4y'' - y = 0\), we have the candidate functions \(y_1(t) = e^{-t/2}\) and \(y_2(t) = -2e^{t/2}\).

• Calculate the first and second derivatives: These derivatives are plugged back into the ODE. For \(y_1(t)\), the second derivative is \(\frac{1}{4}e^{-t/2}\), and for \(y_2(t)\), it is \(\frac{1}{2}e^{t/2}\).
• Substitution and simplification: Upon substituting these derivatives into the main equation, if the ode equals zero, it confirms that the given function is indeed a solution.

Verification gives us confidence that our functions, which could represent physical phenomena, satisfy the governing equations.

Linear combinations

A linear combination of solutions refers to creating a new function from existing solutions by multiplying them with constants and adding them together. This concept is particularly useful when dealing with linear differential equations, such as the given ODE: \(4y'' - y = 0\).

• Expression in terms of solutions: If \(\bar{y}(t)\) is a linear combination of \(y_1(t)\) and \(y_2(t)\), it can be expressed as \(\bar{y}(t) = c_1y_1(t) + c_2y_2(t)\), where \(c_1\) and \(c_2\) are constants.
• Example: For the function \(\bar{y}(t) = 2\cosh(t/2)\), it can be rewritten as \(\bar{y}(t) = \frac{1}{2}y_2(t) - \frac{1}{2}y_1(t)\).

Linear combinations allow the construction of new solutions given that the original solutions are valid. This property holds in linear ODE contexts, providing a broader set of tools for solving differential equations.

Second-order derivatives

Second-order derivatives play a crucial role in the study of second-order differential equations. They describe the rate of change of the rate of change, essentially capturing how the first derivative itself varies over time. In the context of the ODE \(4y'' - y = 0\), it is essential to calculate and correctly apply second-order derivatives to verify solutions.

• Purpose: The second derivative \(y''\) helps in analyzing the concavity or curvature of the function, which can be critical for understanding the behavior of physical systems.
• Calculation: For example, for \(y_1(t) = e^{-t/2}\), the second derivative is \(y_1''(t) = \frac{1}{4}e^{-t/2}\), and for \(y_2(t) = -2e^{t/2}\), it is \(y_2''(t) = \frac{1}{2}e^{t/2}\).

Correct usage of second-order derivatives allows the proper formation and solution of differential equations, as these derivatives are central to the structure of second-order equations.

Hyperbolic functions

Hyperbolic functions, like hyperbolic cosine (\(\cosh\)) and hyperbolic sine (\(\sinh\)), are analogs of regular trigonometric functions but with some unique properties beneficial in differential equations. In our exercise, we used \(\cosh(t/2)\) as part of the solution \(\bar{y}(t)\).

• Definition: \(\cosh(t)\) is defined as \(\frac{e^t + e^{-t}}{2}\), providing a smooth curve similar to \(x^2\) but with exponential growth characteristics.
• Application in ODEs: When combined in expressions, hyperbolic functions can represent a wide range of solution profiles, especially due to their infinite domain and range properties.
• Relation to exponential functions: Since \(\cosh(t)\) can be broken down into an expression involving exponential functions, it is particularly useful in forming solutions in linear differential equations.

Understanding hyperbolic functions enables the solution of differential equations through their linear relations and helpful properties, catering to solutions applicable to real-world phenomena.