Question

Show that \(y(t)=C_{1} e^{2 t}+C_{2} e^{-2 t}\) is a solution of the differential equation \(y^{\prime \prime}-4 y=0\), where \(C_{1}\) and \(C_{2}\) are arbitrary constants.

Short answer

Question: Prove that the function \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\) is a solution to the differential equation \(y'' - 4y = 0\). Answer: We calculated the first and second derivatives of \(y(t)\) and substituted them into the differential equation along with the original function. Upon simplification, we found that the equation holds true, proving that \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\) is a solution to the differential equation \(y'' - 4y = 0\).

Step-by-step solution

Find the first derivative of \(y(t)\)

To find the first derivative, we differentiate \(y(t)\) with respect to \(t\). We must remember to apply the chain rule when differentiating the exponential functions: $$ y'(t) = \frac{d}{dt} [C_1 e^{2t} + C_2 e^{-2t}] $$ $$ y'(t) = C_1 (2e^{2t}) + C_2(-2e^{-2t}) $$

Find the second derivative of \(y(t)\)

Now, differentiate the first derivative \(y'(t)\) with respect to \(t\) again: $$ y''(t) = \frac{d}{dt} [C_1 (2e^{2t}) + C_2(-2e^{-2t})] $$ $$ y''(t) = 4C_1 e^{2t} + 4 C_2 e^{-2t} $$

Substitute the original function and its derivatives into the differential equation

Now, we'll substitute the original function \(y(t)\) and its derivatives \(y'(t)\) and \(y''(t)\) into the differential equation \(y'' - 4y = 0\): $$ (4C_1 e^{2t} + 4 C_2 e^{-2t}) - 4(C_1 e^{2t} + C_2 e^{-2t}) = 0 $$

Simplify the equation and confirm it holds true

Simplify the equation by distributing the \(4\) and combining like terms: $$ 4C_1 e^{2t} + 4 C_2 e^{-2t} - 4C_1 e^{2t} - 4 C_2 e^{-2t} = 0 $$ Notice that the left side of the equation consists of terms that perfectly cancel each other out: $$ \cancel{4C_1 e^{2t}} - \cancel{4C_1 e^{2t}} + \cancel{4 C_2 e^{-2t}} - \cancel{4 C_2 e^{-2t}} = 0 $$ Thus, the left side simplifies to zero, making the equation true: $$ 0 = 0 $$ Since the equation holds true, we have shown that \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\) is a solution to the differential equation \(y'' - 4y = 0\).

Key concepts

Second Order Differential Equation

A second order differential equation involves derivatives up to the second degree of a function. This type of equation is characterized by the appearance of the second derivative, denoted as \(y''\), in the equation. For instance, the equation \(y'' - 4y = 0\) is a typical second order differential equation. These equations are fundamental in modeling various physical phenomena, such as oscillations and wave propagation.

When solving such equations, the main goal is to find a function \(y(t)\) that satisfies the equation. The solution often involves arbitrary constants, like \(C_1\) and \(C_2\), which are determined by initial conditions or boundary values that specify the particular problem. Understanding how these constants adjust solutions to fit specific criteria is crucial in practical applications.

Second order differential equations can usually be solved analytically. Techniques for solving them include the characteristic equation method, which involves finding characteristic roots that simplify the equation into something more manageable.

Solutions to Differential Equations

Finding solutions to differential equations involves identifying a function or a family of functions that satisfy the given equation. This involves substituting the solution back into the original equation and verifying it leads to a true statement. In our case, we have \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\) as a candidate solution.

To verify this solution, we take derivatives of the function and substitute them back into the differential equation. In the example provided, we first found the first derivative \(y'(t)\) and the second derivative \(y''(t)\). Substituting these into \(y'' - 4y = 0\) showed that the terms simplify and cancel out to zero, confirming the solution satisfies the equation.

Different kinds of solutions like general, particular, or specific can emerge based on initial or boundary conditions. Here, \(C_1\) and \(C_2\) are undetermined, making it a general solution. If initial conditions were provided, they would allow solving for these constants, giving a specific solution.

Exponential Functions

Exponential functions, such as \(e^{2t}\) and \(e^{-2t}\), are crucial in solving many differential equations, particularly linear ones with constant coefficients. These functions grow or decay exponentially over time and are defined by the base of the natural logarithm, \(e\).

The primary characteristic of exponential functions is their simplicity in differentiation and integration. Differentiating \(e^{at}\) with respect to \(t\) gives \(ae^{at}\), a form that remains virtually unchanged, making them ideal for expressing solutions to differential equations. This behaviour allows solutions to be expressed as linear combinations, like \(C_1 e^{2t} + C_2 e^{-2t}\), which greatly simplifies solving process.

Exponential solutions often arise in cases where the differential equation models exponential growth or decay scenarios, such as population dynamics, radioactive decay, or in the context of electrical circuits and mechanical vibrations.