Problem 2 - Advanced Engineering Mathematics

Question

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 to the initial-value problem is \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

Step 1: Identify the Given Initial Conditions

You are given the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \). These will be used later to determine the constants \( c_1 \) and \( c_2 \) in the solution.

Step 2: Calculate the First Derivative of y

The general solution is \( y = c_1 e^{4x} + c_2 e^{-x} \). Differentiate this function with respect to \( x \) to find the first derivative:\[ y' = 4c_1 e^{4x} - c_2 e^{-x} \]

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

Substitute \( x = 0 \) into the general solution:\[ y(0) = c_1 e^{0} + c_2 e^{0} = c_1 + c_2 = 1 \]

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

Substitute \( x = 0 \) into the first derivative:\[ y'(0) = 4c_1 e^{0} - c_2 e^{0} = 4c_1 - c_2 = 2 \]

Step 5: Solve the System of Equations

Now solve the two equations from Steps 3 and 4:\[ c_1 + c_2 = 1 \]\[ 4c_1 - c_2 = 2 \]Add the equations to eliminate \( c_2 \):\[ (c_1 + c_2) + (4c_1 - c_2) = 1 + 2 \]\[ 5c_1 = 3 \]\[ c_1 = \frac{3}{5} \]Substitute \( c_1 = \frac{3}{5} \) back into the first equation:\[ \frac{3}{5} + c_2 = 1 \]\[ c_2 = 1 - \frac{3}{5} = \frac{2}{5} \]

Step 6: Write the Particular Solution

Substitute \( c_1 = \frac{3}{5} \) and \( c_2 = \frac{2}{5} \) back into the general solution:\[ y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \]

Step 7: Verify the Particular Solution

To ensure correctness, substitute \( x = 0 \) into the particular solution and its derivative to verify the initial conditions:For \( y(0) = \frac{3}{5} + \frac{2}{5} = 1 \) and \( y'(0) = 4 \times \frac{3}{5} - \frac{2}{5} = 2 \), both conditions are satisfied.

Key Concepts

Differential Equation Solution

A differential equation solution involves finding a function that satisfies the equation, which usually contains derivatives of the function. In this exercise, the differential equation is a second-order linear equation: \[ y'' - 3y' - 4y = 0 \] The solution to this equation is expressed as a combination of exponential functions. These types of solutions are commonly found in differential equations due to their convenient properties when differentiated. Generally, a solution that encompasses all possible specific solutions is termed the 'general solution.' Here, the general solution is: \[ y = c_1 e^{4x} + c_2 e^{-x} \] where \( c_1 \) and \( c_2 \) are constants that will be determined by applying initial conditions. Differential equations often require fitting the solution to specific conditions to be useful in real-world scenarios. Each given initial condition provides one equation, helping us solve for these constants.

General Solution of Differential Equations

The general solution of a differential equation is the overarching solution that contains all specific solutions to that equation. It typically includes arbitrary constants, which need to be determined through additional information, such as initial conditions. In this exercise, the general solution of the differential equation \( y'' - 3y' - 4y = 0 \) involves the exponential solutions given by: \[ y = c_1 e^{4x} + c_2 e^{-x} \] General solutions are vital because they represent the complete set of possibilities for a given differential equation. Once initial conditions, such as \( y(0) = 1 \) and \( y'(0) = 2 \), are specified, a particular solution can be found by plugging these conditions into the general solution and solving for the constants \( c_1 \) and \( c_2 \). This process fits the general pattern to the specific scenario defined by the problem, providing a unique solution that satisfies both the differential equation and the initial conditions.

System of Equations

Solving for the constants in the general solution requires establishing a system of equations. In this context, when initial conditions are applied, they produce a set of algebraic equations that can be solved simultaneously. The initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \) lead to the following system of equations: \[ c_1 + c_2 = 1 \] \[ 4c_1 - c_2 = 2 \] To solve these equations, we can use methods such as substitution or elimination. In the given solution, the elimination method was used, by adding the equations to remove \( c_2 \): - Add both equations: \( (c_1 + c_2) + (4c_1 - c_2) = 1 + 2 \)- Simplifying gives: \( 5c_1 = 3 \), which leads to \( c_1 = \frac{3}{5} \)- Substitute \( c_1 \) back into \( c_1 + c_2 = 1 \) to find \( c_2 = \frac{2}{5} \) This system of equations approach allows deriving constants that define the particular solution, thereby tailoring the general solution to the constraints posed by the initial conditions.