Problem 13 - 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. \(y=c_{1} e^{x} \cos x+c_{2} e^{x} \sin x ; y^{\prime \prime}-2 y^{\prime}+2 y=0\) (a) \(y(0)=1, y^{\prime}(\pi)=0\) (b) \(y(0)=1, y(\pi)=-1\) (c) \(y(0)=1, y(\pi / 2)=1\) (d) \(y(0)=0, y(\pi)=0\)

Step-by-Step Solution

Answer

Solutions exist for cases (a) and (d) only.

Step 1: Understanding the General Solution

The general solution of the differential equation \( y'' - 2y' + 2y = 0 \) is given as \( y = c_1 e^x \cos x + c_2 e^x \sin x \), which involves two arbitrary constants, \( c_1 \) and \( c_2 \).

Step 2: Compute First Derivative

To use the boundary conditions, we first need \( y' \). Differentiate \( y \) with respect to \( x \):\[y' = c_1 (e^x \cos x - e^x \sin x) + c_2 (e^x \sin x + e^x \cos x) = e^x[(c_1 - c_2)\cos x + (c_1 + c_2)\sin x].\]

Step 3: Apply Boundary Conditions for Case (a)

Substitute \( x = 0 \) in \( y \): \[ y(0) = c_1 \cos 0 + c_2 \sin 0 = c_1 = 1. \]Next, substitute \( x = \pi \) in \( y' \) and set the boundary condition \( y'(\pi) = 0 \): \[ y'(\pi) = e^\pi [(-c_1 + c_2)\cos \pi + (c_1 + c_2)\sin \pi] = e^\pi [(-1 + c_2)(-1)] = 0. \]This gives:\( 1-c_2=0 \) or \( c_2 = 1 \).

Step 4: Verify Case (a)

With \( c_1 = 1 \) and \( c_2 = 1 \), the solution satisfies both boundary conditions:\[ y(0) = (c_1 \cdot 1) + (c_2 \cdot 0) = 1,\quad y'(\pi) = e^{\pi}((c_1 - c_2)\cos \pi) = 0. \]A solution exists for condition (a).

Step 5: Apply Boundary Conditions for Case (b)

Substitute \( x = 0 \) in \( y \):\[ y(0) = c_1 = 1. \]Next, substitute \( x = \pi \) in \( y \) for the second boundary condition \( y(\pi) = -1 \):\[ y(\pi) = e^{\pi}[-c_1] = -c_1 e^{\pi} = -1. \]This implies \( c_1 e^{\pi} = 1 \), which is a contradiction since \( c_1 = 1 \).

Step 6: Evaluate Feasibility for Case (b)

The contradiction (\( e^{\pi} eq 1 \)) means no consistent solution exists for boundary condition set (b).

Step 7: Apply Boundary Conditions for Case (c)

Under the boundary conditions for case (c), substitute \( x = 0 \) to get \( y(0) = 1 \) as usual from \( c_1 = 1 \).Then, \( y(\pi/2) = e^{\pi/2}[-c_1 \sin \pi/2 + c_2 \cos \pi/2] = -e^{\pi/2} c_1 = 1 \) cannot be solved with consistent \( c_1 \), due to \(-e^{\pi/2} eq 1 \).

Step 8: Evaluate Feasibility for Case (c)

Case (c) also results in a contradiction similar to case (b), indicating no solution exists.

Step 9: Apply Boundary Conditions for Case (d)

Start by setting \( x = 0 \) in \( y \) to achieve \( y(0) = 0 \), implying \( c_1 = 0 \).Next, assign \( x = \pi \) for \( y(\pi) = 0 \) leading to:\[ y(\pi) = e^{\pi} c_2 \sin \pi = 0. \]

Step 10: Verify Case (d)

The boundary conditions for (d) give consistent results from the equation setup: \( y(0)=0 \) and satisfies \( y(\pi)=0 \) even with arbitrary \( c_2 \), a valid solution exists.

Key Concepts

Boundary Value Problems

Boundary value problems are an essential area in the study of differential equations. These problems aim to find a solution to a differential equation that also satisfies certain specified conditions known as boundary conditions. Boundary conditions specify values that the solution must take on at the boundary of the domain.

In this exercise, we deal with boundary conditions that involve values of the function at certain points, such as at the start and end of an interval. We have examples like setting the values of the function or its derivative at

• the initial point, say \(x = 0\),
• and another point like \(x = \pi\) or \(x = \pi/2\).

Boundary value problems are commonly solved using techniques such as the method of undetermined coefficients or variation of parameters. Successfully addressing these problems helps us understand how solutions behave under constraints that align them with real-world scenarios.

General Solution

The general solution of a differential equation includes all possible solutions of the equation. In this case, the general solution is expressed as a linear combination of two terms, with arbitrary constants \(c_1\) and \(c_2\): \[ y = c_1 e^x \cos x + c_2 e^x \sin x \]

For linear differential equations, the general solution represents a family of functions that contain all specific solutions. Typically, the general solution is derived by solving the characteristic equation associated with the differential equation, resulting in exponential and trigonometric terms. From there, particular solutions that meet given boundary conditions can be extracted by tuning the arbitrary constants.

Our specific task is to find particular instances of this general solution which satisfy individual boundary conditions. This is achieved by substituting the boundary values into the general solution and solving the resulting equations for the constants \(c_1\) and \(c_2\).

Arbitrary Constants

In differential equations, arbitrary constants are placeholders representing a family of solutions. For our given differential equation, the arbitrary constants \(c_1\) and \(c_2\) can be adjusted to match specific boundary conditions.

These constants arise because differential equations involve derivatives, which inherently lose some information about the original function. To recover a unique solution, boundary or initial conditions are used to solve for these constants.Let’s consider

• if the problem specifies \(y(0) = 1\), this helps in finding the value of \(c_1\).
• The second boundary condition may be used to determine \(c_2\).

Upon substituting into the general solution,

• applying one boundary condition helps reduce one arbitrary constant,
• and the additional condition ensures the determination of the other.

Thus, converting the general form to a specific solution involves calculated substitution and adjustment of these constants.

Existence of Solutions

The existence of solutions for a differential equation with boundary conditions concerns whether it's possible to find values for arbitrary constants that satisfy all provided conditions.

In our exercise, we explored different sets of boundary conditions:

• Condition (a) led to a successful determination of values for \(c_1\) and \(c_2\), indicating that a solution exists.
• Conditions (b) and (c) resulted in contradictions, indicating that solutions under these boundary requirements are not possible with the given differential equation.
• For condition (d), the boundary values work harmoniously, allowing a valid solution to exist, even with one arbitrary constant left free.

This exploration demonstrates how the interplay between the type of differential equation and the imposed boundary conditions controls the feasibility of achieving a valid and specific solution. Troubleshooting such contradictions involves reviewing whether the boundary conditions are consistent and suitable for the mathematical construct of the problem.