Solving Boundary Value Problems using the Method of Undetermined Coefficients

In this article, we will explore how to solve a specific boundary value problem using the method of undetermined coefficients. This technique is often employed to find a particular solution to non-homogeneous differential equations, particularly when the non-homogeneous term is a polynomial, exponential, trigonometric, or a combination of these functions. We will provide a detailed step-by-step solution to the given boundary value problem and explain the underlying principles.

The Given Differential Equation

The given differential equation is:

y'' - 2y' - 2y 2x - 2

Characteristic Equation and Homogeneous Solution

The first step in solving a differential equation with the method of undetermined coefficients is to find the corresponding characteristic equation. The characteristic equation for the given differential equation is:

λ^2 - 2λ - 2 0

Solving this quadratic equation, we find the eigenvalues:

λ 1 i, 1 - i

Therefore, the homogeneous solution is:

y_h C_1e^{(1 i)x} C_2e^{(1-i)x}

This can be rewritten using Euler's formula:

y_h C_1e^x[cos(x) isin(x)] C_2e^x[cos(x) - isin(x)]

Simplifying, we get:

y_h e^x[C_1cos(x) iC_1sin(x) C_2cos(x) - iC_2sin(x)]

This further simplifies to:

y_h e^x[(C_1 C_2)cos(x) i(C_1 - C_2)sin(x)]

For simplicity, we denote:

C_1 C_2 A, i(C_1 - C_2) B

Thus, the homogeneous solution becomes:

y_h Ae^xcos(x) Be^xsin(x)

Particular Integral

To find the particular integral, we assume a form that matches the right-hand side of the given differential equation. Since the right-hand side is a linear polynomial, we assume:

yp Ax B

Substituting this into the differential equation and equating coefficients, we solve for A and B. After simplification, we find:

A 1, B 0

Thus, the particular integral is:

yp x

General Solution

The general solution is the sum of the homogeneous and particular solutions:

y y_h yp e^x(Acos(x) Bsin(x)) x

Applying Boundary Conditions

To find the constants A and B, we apply the boundary conditions. The first boundary condition is:

y(0) 0

Substituting x 0 into the general solution, we get:

Acos(0) Bsin(0) 0

This simplifies to:

A 0

Thus, the solution simplifies to:

y Be^xsin(x) x

The second boundary condition is:

y(π) π

Substituting x π into the simplified solution, we get:

Be^πsin(π) π π

Since sin(π) 0, we have:

π π

This condition is satisfied for any B, but we need a specific solution. Let's denote:

B e^{-π}

Therefore, the final solution is:

y e^{-π}e^xsin(x) x

Or, more concisely:

y e^{x-π}sin(x) x

Real Solutions

If we are looking for real solutions, we can set A as purely imaginary. Thus, the solution becomes:

y Ce^xsin(x) x

Where C is a real constant.

Conclusion

By following the method of undetermined coefficients, we were able to solve the given boundary value problem and find the general solution. Through the application of boundary conditions, we determined specific constants for the solution. This detailed process can be applied to other similar boundary value problems involving polynomials and exponential functions.

Keywords and Relevant Terms

boundary value problem, method of undetermined coefficients, differential equation