Solving Second-Order Linear Non-Homogeneous Differential Equations with the Method of Undetermined Coefficients

In this article, we will explore how to solve a specific type of second-order linear non-homogeneous differential equation using the method of undetermined coefficients. Specifically, we will solve the equation:

1. Problem Definition and Solution Outline

Consider the differential equation:

y'' - 4y' - 5y e^{-2x} cos x

The solution to this equation will consist of two parts: the complementary solution y_c) and the particular solution y_p).

2. Finding the Complementary Solution (y_c)

To find the complementary solution, we first solve the related homogeneous equation:

y'' - 4y' - 5y 0

and determine its roots using the characteristic equation:

r^2 - 4r - 5 0

Solving this quadratic equation, we get:

r frac{-b pm sqrt{b^2 - 4ac}}{2a} frac{4 pm sqrt{16 20}}{2} frac{4 pm sqrt{36}}{2} 2 pm 3

This gives us the roots (r 5) and (r -1).

3. Finding the Particular Solution (y_p)

Next, we find a particular solution (y_p) for the non-homogeneous equation. Given the right-hand side is (e^{-2x} cos x), we need to use the method of undetermined coefficients.

Since (e^{-2x} cos x) is already present in the complementary solution (as a complex solution (e^{-2x} (cos x i sin x)) breaks down to (e^{-2x} cos x) and (e^{-2x} sin x)), we must modify our guess to include an extra factor of (x).

We assume a particular solution of the form:

y_p x e^{-2x} (A cos x B sin x)

4. Computing Derivatives

Now, we need to compute the first and second derivatives of (y_p).

First Derivative (y_p')

Using the product rule:

y_p' e^{-2x} (A cos x B sin x) - 2x e^{-2x} (A cos x B sin x) x e^{-2x} (-2A sin x 2B cos x)

Second Derivative (y_p'')

This will involve applying the product rule again and simplifying.

5. Substituting into the Differential Equation

Substitute (y_p), (y_p'), and (y_p'') back into the left side of the original differential equation:

y'' - 4y' - 5y e^{-2x} cos x

6. Collecting Like Terms

Set the coefficients of (e^{-2x} cos x) and (e^{-2x} sin x) on both sides of the equation equal to each other to solve for the constants (A) and (B).

7. Writing the General Solution

The general solution of the original differential equation is:

y y_c y_p e^{-2x} C_1 cos x C_2 sin x x e^{-2x} (A cos x B sin x)

Where (C_1) and (C_2) are constants determined from initial or boundary conditions.

Conclusion

This process outlines the method for solving the given differential equation. To find the explicit values of (A) and (B), you need to perform the necessary derivative calculations and algebra. If you need further assistance with the calculations or have specific values to use, feel free to ask!

Related Keywords

Differential Equations Undetermined Coefficients Method Characteristic Equations