Problem 11 - Advanced Engineering Mathematics

Question

In Problems \(1-18\), solve each differential equation by variation of parameters. $$ y^{\prime \prime}+3 y^{\prime}+2 y=\frac{1}{1+e^{x}} $$

Step-by-Step Solution

Answer

The general solution is found by combining the complementary solution and the particular solution obtained through variation of parameters.

Step 1: Find the Complementary Solution

First, find the roots of the characteristic equation associated with the homogeneous part of the differential equation. The characteristic equation is \( r^2 + 3r + 2 = 0 \). Solve this quadratic equation using the formula \( r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Substitute \( a = 1 \), \( b = 3 \), \( c = 2 \) to find the roots. This gives us \( r = -1 \) and \( r = -2 \). Therefore, the complementary solution is \( y_c = C_1 e^{-x} + C_2 e^{-2x} \).

Step 2: Apply Variation of Parameters

For variation of parameters, we express the particular solution as \( y_p = u_1 y_1 + u_2 y_2 \), where \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \). The functions \( u_1 \) and \( u_2 \) are solutions of the system: \( u_1' y_1 + u_2' y_2 = 0 \) and \( u_1' y_1' + u_2' y_2' = \frac{1}{1+e^x} \).

Step 3: Set up the Wronskian and Solve for u1' and u2'

Compute the Wronskian of \( y_1 \) and \( y_2 \): \[ W = \begin{vmatrix} e^{-x} & e^{-2x} \ -e^{-x} & -2e^{-2x} \end{vmatrix} = -e^{-3x}. \]Use the system equations to solve for \( u_1' \) and \( u_2' \):1. \( u_1' e^{-x} + u_2' e^{-2x} = 0 \)2. \( u_1' (-e^{-x}) + u_2' (-2e^{-2x}) = \frac{1}{1+e^x} \). By solving this, find:\( u_1' = -\frac{e^{-x}}{1+e^x} \) and \( u_2' = \frac{2e^{-2x}}{1+e^x} \).

Step 4: Integrate u1' and u2' to Find u1 and u2

Integrate \( u_1' \) and \( u_2' \) to find \( u_1 \) and \( u_2 \):\( u_1 = \int -\frac{e^{-x}}{1+e^x} \, dx \) and \( u_2 = \int \frac{2e^{-2x}}{1+e^x} \, dx \).These integrals can be solved using substitution:- \( u_1 = - \ln |1+e^x| \)- \( u_2 = \text{(solve using integration by parts or a suitable substitution)} \)

Step 5: Form the Particular Solution

Substitute \( u_1 \) and \( u_2 \) back into \( y_p = u_1 y_1 + u_2 y_2 \). Compute to find the particular solution. Combine it with the complementary solution to form the general solution.

Step 6: Write Down the General Solution

The general solution of the differential equation is \( y = y_c + y_p \). Substitute \( y_c \) and \( y_p \) with their respective expressions obtained from the previous steps to write down the solution.

Key Concepts

Wronskian determinant

The Wronskian determinant is a mathematical tool used to determine whether a set of solutions forms a fundamental set of solutions for a linear differential equation. If you're dealing with two solutions, say \( y_1 \) and \( y_2 \), the Wronskian is constructed by arranging these solutions and their derivatives into a 2x2 matrix.

• For our problem, the solutions are \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \).

To compute the Wronskian, we form the matrix:\[W = \begin{vmatrix} e^{-x} & e^{-2x} \ -e^{-x} & -2e^{-2x} \end{vmatrix}\]Calculating the determinant, we get:\[ W = e^{-x}(-2e^{-2x}) - (e^{-2x}(-e^{-x})) = -e^{-3x} \]This non-zero result indicates that \( y_1 \) and \( y_2 \) are linearly independent, hence suitable for these solutions. Linearly independent functions are essential, as they ensure the complete system of solutions for our differential equation.

Complementary Solution

The complementary solution to a differential equation represents the solution to the homogeneous part of the equation. This is typically the first step in solving non-homogeneous differential equations, like the one we have here.

• It involves solving the characteristic equation derived from the differential equation's homogeneous form.
• For an equation \( y'' + 3y' + 2y = 0 \), the characteristic equation is \( r^2 + 3r + 2 = 0 \).

Solving this quadratic equation using the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By substituting \( a = 1 \), \( b = 3 \), and \( c = 2 \), we find the roots:\( r = -1 \) and \( r = -2 \).Thus, our complementary solution is:\[ y_c = C_1 e^{-x} + C_2 e^{-2x} \]This expression represents the general solution to the equation without any external forcing function, such as the \( \frac{1}{1+e^x} \) present in this problem.

Particular Solution

The particular solution is the solution to the non-homogeneous part of the differential equation, which deals with the external term. In our problem, the non-homogeneous term is \( \frac{1}{1+e^x} \).To find the particular solution \( y_p \) using variation of parameters:

• We express it as \( y_p = u_1 y_1 + u_2 y_2 \), where \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \).
• The functions \( u_1 \) and \( u_2 \) need to satisfy the system of equations: \( u_1' y_1 + u_2' y_2 = 0 \) and \( u_1' y_1' + u_2' y_2' = \frac{1}{1+e^x} \).

Using the Wronskian, solve for \( u_1' \) and \( u_2' \):- \( u_1' = -\frac{e^{-x}}{1+e^x} \)- \( u_2' = \frac{2e^{-2x}}{1+e^x} \)Next, integrate these to find \( u_1 \) and \( u_2 \):- \( u_1 = - \ln |1+e^x| \) (using a suitable substitution)Integrate \( u_2' \) similarly to find \( u_2 \). Finally, substitute \( u_1 \) and \( u_2 \) back into \( y_p \) to get the specific expression for the non-homogeneous solution.The complete solution will be the sum of the complementary and particular solutions: \( y = y_c + y_p \). This gives us the general solution to the original differential equation.