Problem 66 - Calculus Volume 3

Question

a. Write the form for the particular solution \(y_{p}(x)\) for the method of undetermined coefficients. b. Use a computer algebra system to find a particular solution to the given equation. $$ y^{\prime \prime}-y^{\prime}-y=x+e^{-x} $$

Step-by-Step Solution

Answer

The particular solution is \( y_p(x) = -x + e^{-x} \).

Step 1: Write the characteristic equation

The first step is to find the characteristic equation of the homogeneous part of the differential equation. The homogeneous equation is: \[ y'' - y' - y = 0 \]The characteristic equation associated with this is found by assuming a solution of the form \( y = e^{rx} \), leading to:\[ r^2 - r - 1 = 0 \]

Step 2: Solve the characteristic equation

Solve the characteristic equation \( r^2 - r - 1 = 0 \) using the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -1 \), and \( c = -1 \).This gives:\[ r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]The roots are \( r_1 = \frac{1 + \sqrt{5}}{2} \) and \( r_2 = \frac{1 - \sqrt{5}}{2} \).

Step 3: Construct the homogeneous solution

Using the roots from Step 2, the general solution of the homogeneous equation is:\[ y_h = C_1 e^{\frac{1 + \sqrt{5}}{2}x} + C_2 e^{\frac{1 - \sqrt{5}}{2}x} \]

Step 4: Identify the form of particular solution

For the method of undetermined coefficients, the non-homogeneous part \( x + e^{-x} \) suggests choosing a particular solution of the form:\[ y_p(x) = Ax + B + Ce^{-x} \]This is based on polynomial terms for \( x \) and exponential terms for \( e^{-x} \).

Step 5: Find the derivatives of the particular solution

Calculate the derivatives of the particular solution \( y_p(x) = Ax + B + Ce^{-x} \):\[ y_p'(x) = A - Ce^{-x} \]\[ y_p''(x) = Ce^{-x} \]

Step 6: Substitute into the differential equation

Substitute back into the differential equation:\[ Ce^{-x} - (A - Ce^{-x}) - (Ax + B + Ce^{-x}) = x + e^{-x} \]Simplify the expressions:\[ 0 - A - Ax - B = x + e^{-x} \]This gives the equation to solve for \( A \), \( B \), and \( C \).

Step 7: Equate coefficients and solve for constants

Comparing coefficients from each side results in a system of equations:For the \( e^{-x} \):-\( C = 1 \) For the \( x \):-\( A = -1 \) For the constant term:-\( -B = 0 \) From which we find \( C = 1 \), \( A = -1 \), and \( B = 0 \).

Step 8: Write the particular solution

The particular solution is:\[ y_p(x) = -x + e^{-x} \]

Step 9: Finalize the particular solution using a CAS

After confirmation with a computer algebra system (CAS) the particular solution remains valid and confirms the manual calculation. Therefore, the particular solution is:\[ y_p(x) = -x + e^{-x} \]

Key Concepts

Characteristic Equation

When solving linear differential equations, a critical first step is finding the characteristic equation. This equation comes from the homogeneous part of the differential equation, which means looking at the equation without any non-homogeneous (or external) components. For our given differential equation, \[ y'' - y' - y = 0, \]we assume a solution of the form \( y = e^{rx} \). By substituting this into the homogeneous equation, we get a polynomial known as the characteristic equation:\[ r^2 - r - 1 = 0. \]This quadratic equation is key to determining the types of solutions (or roots) that will define the behavior of the differential equation's solutions. To solve it, we apply the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \( a = 1 \), \( b = -1 \), and \( c = -1 \). Solving this gives characteristic roots \( r_1 \) and \( r_2 \), which are used in creating the homogeneous solution.

Homogeneous Solution

The roots found from the characteristic equation tell us the nature of the homogeneous solution. In our example, the characteristic roots are:\[ r_1 = \frac{1 + \sqrt{5}}{2} \] and \[ r_2 = \frac{1 - \sqrt{5}}{2}. \]Each root represents an exponential function that forms part of the homogeneous solution. Thus, the general form of the homogeneous solution is:\[ y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x}, \]where \( C_1 \) and \( C_2 \) are constants determined by initial conditions. These constants shape the solution to match particular conditions given in any practical problem. In this context, the homogeneous solution represents the underlying behavior without any external forces or influences captured by the non-homogeneous part. The homogeneous solution describes the "natural response" of the system defined by the differential equation.

Particular Solution

To find the particular solution, we use the method of undetermined coefficients, which works well when the non-homogeneous part of the differential equation is a combination of polynomials, exponentials, or sines and cosines. In our differential equation:\[ y'' - y' - y = x + e^{-x}, \]we have non-homogeneous terms: \( x \) and \( e^{-x} \). These suggest guessing a form for the particular solution:\[ y_p(x) = Ax + B + Ce^{-x}. \]Here:

• \( Ax + B \) accounts for the polynomial \( x \), and
• \( Ce^{-x} \) accounts for the exponential \( e^{-x} \).

We then differentiate this assumed form to find its first and second derivatives, and substitute them back into the original differential equation. By equating coefficients on both sides of the equation, we solve for the unknown coefficients \( A \), \( B \), and \( C \). This process yields:

• \( A = -1 \),
• \( B = 0 \),
• \( C = 1 \).

Finally, substituting these values back into the guessed form gives the particular solution:\[ y_p(x) = -x + e^{-x}. \]This solution encompasses the influence of external inputs or forces represented by the non-homogeneous terms in the differential equation.