Problem 14 - Advanced Engineering Mathematics

Question

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 (c) with \(c_1+c_2=-3\) and (d) with \(c_1=-1, c_2=1\). No solutions for (a) or (b).

Step 1: Find the First and Second Derivatives of y

Given the family of solutions, \(y = c_1 x^2 + c_2 x^4 + 3\), we first find the first and second derivatives of \(y\). First derivative: \(y' = 2c_1 x + 4c_2 x^3\)Second derivative: \(y'' = 2c_1 + 12c_2 x^2\).

Step 2: Substitute into the Differential Equation

Substitute the expressions for \(y\), \(y'\), and \(y''\) into the given differential equation: \(x^2 y'' - 5xy' + 8y = 24\).By substituting the derivatives:\(x^2(2c_1 + 12c_2 x^2) - 5x(2c_1 x + 4c_2 x^3) + 8(c_1 x^2 + c_2 x^4 + 3) = 24\).

Step 3: Simplify the Differential Equation

Simplify the equation from the previous step:\(2c_1 x^2 + 12c_2 x^4 - 10c_1 x^2 - 20c_2 x^4 + 8c_1 x^2 + 8c_2 x^4 + 24 = 24\).Combine like terms:\(0x^4 + 0x^2 + 24 = 24\).The equation simplifies to a true statement, verifying that the function is indeed a solution to the differential equation.

Step 4 (a): Check Boundary Conditions y(-1)=0, y(1)=4

For \(y(-1)=0\), we substitute into the solution: \(c_1 (-1)^2 + c_2 (-1)^4 + 3 = 0\) leads to \(c_1 + c_2 = -3\).For \(y(1)=4\), substitute:\(c_1 (1)^2 + c_2 (1)^4 + 3 = 4\) leads to \(c_1 + c_2 + 3 = 4\), which simplifies to \(c_1 + c_2 = 1\).The two equations \(c_1+c_2=-3\) and \(c_1+c_2=1\) are inconsistent, indicating no such member exists.

Step 4 (b): Check Boundary Conditions y(0)=1, y(1)=2

For \(y(0)=1\), substitute into the solution:\(c_1 (0)^2 + c_2 (0)^4 + 3 = 1\) simplifies to 3 = 1, which is inconsistent.This boundary condition is impossible to satisfy, so no member exists in this family.

Step 4 (c): Check Boundary Conditions y(0)=3, y(1)=0

For \(y(0)=3\), we have:\(c_1 (0)^2 + c_2 (0)^4 + 3 = 3\), satisfied for any \(c_1, c_2\).For \(y(1)=0\):\(c_1 (1)^2 + c_2 (1)^4 + 3 = 0\), which simplifies to \(c_1 + c_2 = -3\).Thus, a solution \(c_1 + c_2 = -3\) satisfies both conditions.

Step 4 (d): Check Boundary Conditions y(1)=3, y(2)=15

For \(y(1)=3\), substitute:\(c_1 (1)^2 + c_2 (1)^4 + 3 = 3\), leading to \(c_1 + c_2 = 0\).For \(y(2)=15\), substitute:\(c_1 (2)^2 + c_2 (2)^4 + 3 = 15\) gives \(4c_1 + 16c_2 = 12\), simplifying to \(c_1 + 4c_2 = 3\).The system of equations \(c_1 + c_2 = 0\) and \(c_1 + 4c_2 = 3\) can be solved to find values of \(c_1\) and \(c_2\). Substituting \(c_1 = -c_2\) into the second equation gives \(-c_2 + 4c_2 = 3\), leading to \(3c_2 = 3\) and \(c_2 = 1\). Then \(c_1 = -1\). Both conditions are satisfied.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. In these equations, derivatives represent change and play a crucial role in describing various physical phenomena. In our exercise, understanding the given differential equation:\[x^2 y^{\prime\prime} - 5x y^{\prime} + 8y = 24\]is key. The equation relates a function \(y\) to its first (\(y'\) ) and second derivatives (\(y''\)). Such equations are frequently used to model situations ranging from motion to heat flow.
The main goal when dealing with differential equations is to find a function \(y\), or a family of functions, that satisfies the particular equation. In this specific problem, the function involves parameters \(c_1\) and \(c_2\), making it a two-parameter family.

Solution Verification

Solution verification ensures that the proposed function actually satisfies the given differential equation. For this problem, a family of functions \(y = c_1 x^2 + c_2 x^4 + 3\) is provided as a solution. The process involves calculating the first and second derivatives:

• First derivative: \(y' = 2c_1 x + 4c_2 x^3\)
• Second derivative: \(y'' = 2c_1 + 12c_2 x^2\)

Substituting these into the differential equation helps verify if the family fully satisfies it. The simplified calculation shows that the equation holds true with these functional forms, meaning the proposed family can indeed be a solution. This step is crucial before any further analysis, such as checking boundary conditions.

Two-Parameter Family

A two-parameter family of solutions consists of functions characterized by two arbitrary constants, in this case, \(c_1\) and \(c_2\). This kind of general representation is powerful because it includes a wide range of possible solutions. Our example is:\[y = c_1 x^2 + c_2 x^4 + 3\]Each pair \((c_1, c_2)\) produces a different specific solution within the family. Not all of these pairs will meet the specific requirements, or boundary conditions, imposed by the problem, but by examining them, we can find if any fits. Here, the exercise asks us to determine which members satisfy different boundary conditions, showing the flexibility and breadth of two-parameter families.

Boundary Conditions

Boundary conditions specify the values of a function at particular points, helping to narrow down the exact solution from a family of potential solutions. They are key in determining the specific parameters that fulfill the conditions set out in a problem. In this exercise, four sets of boundary conditions are examined:

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

The task involves substituting these conditions into the solution equation to derive values for \(c_1\) and \(c_2\). In cases (a) and (b), the conditions provide inconsistent or impossible results, showing that there are no valid members in the family for those scenarios. However, conditions (c) and (d) offer consistent equations, and solutions are found by solving these equations for \(c_1\) and \(c_2\), showing how boundary conditions function to pinpoint specific solutions.