Problem 14 - Advanced Engineering Mathematics

Question

In Problems 13 and 14, the given two-parameter family is a solution of the indicated differential equation on the interval \((-\infty, \infty)\). Determine whether a member of the family can be found that satisfies the boundary conditions. $$ y=c_{1} x^{2}+c_{2} x^{4}+3 ; x^{2} y^{\prime \prime}-5 x y^{\prime}+8 y=24 $$ (a) \(y(-1)=0, y(1)=4\) (b) \(y(0)=1, y(1)=2\) (c) \(y(0)=3, y(1)=0\) (d) \(y(1)=3, y(2)=15\)

Step-by-Step Solution

Answer

Solutions exist for boundary conditions (c) and (d).

Step 1: Understand the Given Problem

We are given a two-parameter family function \( y = c_1 x^2 + c_2 x^4 + 3 \) that is a solution to the differential equation \( x^2 y'' - 5x y' + 8y = 24 \). We need to determine parameters \( c_1 \) and \( c_2 \) to satisfy given boundary conditions.

Step 2: Calculate the Derivatives of the Family Function

The first derivative of \( y \) is \( y' = 2c_1x + 4c_2x^3 \). The second derivative is \( y'' = 2c_1 + 12c_2x^2 \).

Step 3: Check Boundary Condition (a)

Substitute \( y(-1) = 0 \) and \( y(1) = 4 \) into the family function: \[ \begin{align*}-1: & \quad c_1 (-1)^2 + c_2 (-1)^4 + 3 = 0 \Rightarrow c_1 + c_2 + 3 = 0, \1: & \quad c_1 (1)^2 + c_2 (1)^4 + 3 = 4 \Rightarrow c_1 + c_2 + 3 = 4.\end{align*} \] Both result in contradictions, so no solution is possible for (a).

Step 4: Check Boundary Condition (b)

Substitute \( y(0) = 1 \) and \( y(1) = 2 \) into the family function: \[ \begin{align*}0: & \quad 0 + 0 + 3 = 1 \Rightarrow 3 = 1, \quad ext{(Clearly false)}, \1: & \quad c_1 + c_2 + 3 = 2.\end{align*} \] This also results in contradictions, therefore no solution satisfies (b).

Step 5: Check Boundary Condition (c)

Substitute \( y(0) = 3 \) and \( y(1) = 0 \) into the family function: \[ \begin{align*}0: & \quad 0 + 0 + 3 = 3 \Rightarrow ext{True}, \1: & \quad c_1 + c_2 + 3 = 0 \Rightarrow c_1 + c_2 = -3.\end{align*} \] The first condition holds, so we solve \( c_1 + c_2 = -3 \), which is possible by assigning values like \( c_1 = -4 \), \( c_2 = 1 \). So, a member of the family satisfies (c).

Step 6: Check Boundary Condition (d)

Substitute \( y(1) = 3 \) and \( y(2) = 15 \) into the family function: \[ \begin{align*}1: & \quad c_1 + c_2 + 3 = 3 \Rightarrow c_1 + c_2 = 0, \2: & \quad 4c_1 + 16c_2 + 3 = 15 \Rightarrow 4c_1 + 16c_2 = 12.\end{align*} \] Solving these, assume \( c_1 = -c_2 \); thus, the system is consistent, and a member of the family satisfies (d) particularly \( c_2 = 1, c_1 = -1 \).

Key Concepts

Boundary Conditions

Boundary conditions are essential in determining specific solutions to differential equations. They define the values of the function or its derivatives at particular points, which in turn help specify the unique solution of the differential equation from a general family of solutions.

In the given exercise, the boundary conditions are stated for each scenario, such as:

• Condition (a): \(y(-1)=0, y(1)=4\)
• Condition (b): \(y(0)=1, y(1)=2\)
• Condition (c): \(y(0)=3, y(1)=0\)
• Condition (d): \(y(1)=3, y(2)=15\)

To find a specific member of the family of solutions satisfying the differential equation, these conditions must be applied and tested one by one. When substituted into the equation, they guide us in finding or eliminating particular solution possibilities.

Two-Parameter Family

A two-parameter family of functions involves a solution that depends on two independent parameters, typically denoted as \(c_1\) and \(c_2\). This form allows flexibility in finding many potential solutions by varying these parameters.

In the example, the two-parameter family is expressed as:

• \(y = c_1 x^2 + c_2 x^4 + 3\)

Here, \(c_1\) and \(c_2\) represent two constants that adjust the shape of the parabola within its solution space. Through boundary conditions, specific values for these parameters can be determined to meet the requirements of the differential equation. By adjusting \(c_1\) and \(c_2\), we are essentially fine-tuning which member of this family "fits" the boundary conditions outlined.

Derivatives

Derivatives play a crucial role in analyzing the behavior of functions within differential equations. They provide information on the rate of change of functions, which is necessary to apply and solve differential equations.

For this exercise, the solution involves calculating both the first and second derivatives of the function:

• First derivative: \(y' = 2c_1x + 4c_2x^3\)
• Second derivative: \(y'' = 2c_1 + 12c_2x^2\)

These derivatives are then substituted back into the original differential equation \(x^2 y'' - 5x y' + 8y = 24\). This process shows how changes in \(x\) affect both the slope of \(y\) and its curvature, ultimately helping to map which function graph aligns with the given equation and boundary conditions.

Solution Methods

Solution methods in differential equations are techniques to find functions that satisfy a given differential equation along with any specified conditions.

In this case, the solution method involves substituting the family function and its derivatives into the differential equation. Following this, the boundary conditions are incorporated:

• Solve for the constants in different cases of boundary conditions to check if they can satisfy the equation.

For condition (c), where \(y(0)=3, y(1)=0\), we found specific values of \(c_1 = -4\) and \(c_2 = 1\) that satisfy the boundary expectations. However, for other conditions, no values for \(c_1\) and \(c_2\) satisfy the boundary, showing no solution possible in those scenarios. Such methods and logical trials help narrow down potential solutions and verify their correctness against the differential equation.