Problem 2 - Advanced Engineering Mathematics

Question

In Problems \(1-4\), the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. $$ \begin{aligned} &y=c_{1} e^{4 x}+c_{2} e^{-x},(-\infty, \infty) ; y^{\prime \prime}-3 y^{\prime}-4 y=0, y(0)=1 \\ &y^{\prime}(0)=2 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

Step 1: Understand the General Solution

The given family of functions is the general solution: \( y = c_{1} e^{4x} + c_{2} e^{-x} \). This is derived from the differential equation \( y'' - 3y' - 4y = 0 \). Our task is to find specific constants \( c_1 \) and \( c_2 \) such that this function satisfies the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \).

Step 2: Find the First Derivative

Compute the first derivative of the general solution. Given: \( y = c_{1} e^{4x} + c_{2} e^{-x} \). Using differentiation: \( y' = \frac{d}{dx}(c_{1} e^{4x}) + \frac{d}{dx}(c_{2} e^{-x}) = 4c_{1} e^{4x} - c_{2} e^{-x} \).

Step 3: Apply Initial Condition y(0)=1

Substitute \( x = 0 \) into the general solution \( y(0) = c_{1} e^{4(0)} + c_{2} e^{-0} = c_{1} + c_{2} = 1 \). This gives us the first equation: \( c_1 + c_2 = 1 \).

Step 4: Apply Initial Condition y'(0)=2

Substitute \( x = 0 \) into the derivative: \( y'(0) = 4c_{1} e^{4(0)} - c_{2} e^{-0} = 4c_{1} - c_{2} = 2 \). This gives us the second equation: \( 4c_1 - c_2 = 2 \).

Step 5: Solve the System of Equations

We have the system: 1. \( c_1 + c_2 = 1 \).2. \( 4c_1 - c_2 = 2 \).Add the two equations to eliminate \( c_2 \): \( (c_1 + c_2) + (4c_1 - c_2) = 1 + 2 \), which simplifies to \( 5c_1 = 3 \), leading to \( c_1 = \frac{3}{5} \). Insert \( c_1 \) back into \( c_1 + c_2 = 1 \), giving \( \frac{3}{5} + c_2 = 1 \), thus \( c_2 = \frac{2}{5} \).

Step 6: Write the Particular Solution

With calculated values of \( c_1 \) and \( c_2 \), the particular solution of the initial-value problem is: \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation problem where you are given additional information called initial conditions. This information helps define a unique solution from a family of possible solutions. The initial conditions typically specify the value of the function and its derivatives at a particular point.
In our example, the initial value problem involves the differential equation \(y'' - 3y' - 4y = 0\) with two initial conditions:

• \(y(0) = 1\)
• \(y'(0) = 2\)

These conditions are crucial because they allow us to determine the specific constants in the general solution, thus finding the unique solution that fits them.

General Solution

The general solution of a differential equation contains all possible solutions and is expressed in terms of arbitrary constants. For a second-order linear differential equation like the one given \(y'' - 3y' - 4y = 0\), the general solution is:

• \(y = c_{1} e^{4x} + c_{2} e^{-x}\)

Here, \(c_{1}\) and \(c_{2}\) are arbitrary constants that represent a family of solutions. To find a particular solution from this family, we use initial conditions to determine specific values for these constants. This converts the general solution into a specific one that satisfies the given initial conditions.

System of Equations

A system of equations comes into play when we have multiple conditions or equations that we need to solve together. In this case, applying the initial conditions to our functions and derivatives creates a system of linear equations:

• \(c_1 + c_2 = 1\)
• \(4c_1 - c_2 = 2\)

These equations allow us to solve for \(c_1\) and \(c_2\) by eliminating one variable in favor of the other. Solving these equations gives us the precise constants \(c_1 = \frac{3}{5}\) and \(c_2 = \frac{2}{5}\). By resolving these constants, we arrive at the particular solution to the initial value problem.

Exponential Functions

Exponential functions are a type of mathematical function that involve an exponent and are expressed in the form \(e^{x}\), where \(e\) is the base of the natural logarithm. In the context of solving differential equations, exponential functions often represent solutions because they naturally describe rates of change and behavior of systems over time.
The general solution in the exercise involves exponential terms \(e^{4x}\) and \(e^{-x}\). Different constants \(c_1\) and \(c_2\) modify the influence of these exponential terms on the solution. This shows how exponential functions contribute to the shape and behavior of the solution over the interval \((-\infty, \infty)\). Such functions are crucial as they can model growth, decay, oscillations, and other dynamic behaviors observed in various fields like physics, biology, and economics.