Problem 30 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. \(y^{\prime \prime}+4 y^{\prime}+4 y=(3+x) e^{-2 x}, y(0)=2, y^{\prime}(0)=5\)

Step-by-Step Solution

Answer

The solution is \( y(x) = 2e^{-2x} + 8x e^{-2x} \).

Step 1: Solve the Homogeneous Equation

First, solve the homogeneous equation associated with the given differential equation: \[ y'' + 4y' + 4y = 0. \]To find the characteristic equation, replace \(y''\), \(y'\), and \(y\) with \(r^2\), \(r\), and 1 respectively which gives:\[ r^2 + 4r + 4 = 0. \]This factors as \[ (r+2)^2 = 0. \]Thus, the repeated root is \(r = -2\). The general solution to the homogeneous equation is:\[ y_h(x) = C_1 e^{-2x} + C_2 x e^{-2x}. \]

Step 2: Find Particular Solution

Next, find a particular solution, \( y_p \), to the non-homogeneous differential equation.Since the right-hand side is \((3+x) e^{-2x}\), we choose a trial solution for \(y_p\) of the form:\[ y_p = (Ax^2 + Bx) e^{-2x}. \]Differentiate \(y_p\) to find \(y_p'\) and \(y_p''\), then substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original differential equation and solve for \(A\) and \(B\). Upon solving, you find:\[ A = 0, \quad B = -\frac{3}{2}. \]Hence, the particular solution is \[ y_p(x) = -\frac{3}{2}x e^{-2x}. \]

Step 3: Write the General Solution

Combine the homogeneous solution and the particular solution to form the general solution:\[ y(x) = C_1 e^{-2x} + C_2 x e^{-2x} - \frac{3}{2} x e^{-2x}. \]

Step 4: Apply Initial Conditions

Use the initial conditions \(y(0) = 2\) and \(y'(0) = 5\) to solve for constants \(C_1\) and \(C_2\).Start with \(y(0) = 2\):\[ y(0) = C_1 + 0 - 0 = C_1 = 2. \]Then, find \(y'(0) = 5\).Differentiate the general solution for \(y'(x)\) and substitute \(x = 0\):\[ y'(x) = -2C_1 e^{-2x} - 2C_2 e^{-2x} + C_2 xe^{-2x} - \frac{3}{2} e^{-2x} - 3xe^{-2x}. \]Evaluating at \(x=0\):\[ y'(0) = -2 \cdot 2 - \frac{3}{2} + C_2 = 5. \]This simplifies to:\[ -4 - \frac{3}{2} + C_2 = 5. \]Thus,\[ C_2 = 5 + 4 + \frac{3}{2} = \frac{19}{2}. \]

Step 5: Write the Final Solution

Substitute \(C_1 = 2\) and \(C_2 = \frac{19}{2}\) into the general solution:\[ y(x) = 2e^{-2x} + \frac{19}{2}x e^{-2x} - \frac{3}{2} x e^{-2x}. \]Simplify to get the final solution:\[ y(x) = 2e^{-2x} + 8x e^{-2x}. \]

Key Concepts

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with initial conditions. These conditions specify the value of the function and its derivatives at a certain point. This enables us to find a unique solution to the differential equation. In our example, the differential equation is:

• \( y'' + 4y' + 4y = (3+x) e^{-2x} \)

The initial conditions provided are:

• \( y(0) = 2 \)
• \( y'(0) = 5 \)

These tell us the function's value and the slope of the graph at \(x = 0\). By solving the initial-value problem, we ensure that the found function meets these specific conditions, resulting in a single precise solution to the problem, rather than a general one.

Homogeneous Equation

Homogeneous equations are an important concept in differential equations. They're called homogeneous because the expression is set to zero. For the exercise, the homogeneous equation derived is:

• \( y'' + 4y' + 4y = 0 \)

To solve this, we find the characteristic equation by assuming a solution of the form \( y = e^{rx} \). This substitution leads to the characteristic equation:

• \( r^2 + 4r + 4 = 0 \)

This further factors to:

• \( (r+2)^2 = 0 \)

With a repeated root \( r = -2 \), we construct the general solution for the homogeneous equation:

• \( y_h(x) = C_1 e^{-2x} + C_2 x e^{-2x} \)

This solution utilizes both exponential and polynomial terms due to the repeated root, adapting to handle the multiplicity.

Particular Solution

In solving non-homogeneous differential equations, finding a particular solution is key. It specifically deals with the non-zero right-hand side of the equation. The aim is to find a function that satisfies:

• \( y'' + 4y' + 4y = (3+x) e^{-2x} \)

A trial solution is employed, which often follows the form of the right-hand side of the equation, allowing us to test different polynomials. In our task, the trial solution is chosen as:

• \( y_p = (Ax^2 + Bx) e^{-2x} \)

Through substitution into the original differential equation and equating coefficients, the values of \(A\) and \(B\) are solved, leading to the particular solution:

• \( y_p(x) = -\frac{3}{2}x e^{-2x} \)

Combining this with the homogeneous solution yields the complete solution to the differential equation.

Characteristic Equation

The characteristic equation is fundamental for analyzing and solving linear homogeneous differential equations. It is derived from replacing the differential parts with algebraic expressions. In the example, consider the original homogeneous equation:

• \( y'' + 4y' + 4y = 0 \)

By substituting \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1, we form the characteristic equation:

• \( r^2 + 4r + 4 = 0 \)

The solutions to this quadratic equation provide the values of \( r \) that are used to construct the general form of the solution. In our case:

• The equation factors to \( (r+2)^2 = 0 \), yielding a double root \( r = -2 \).

This spacing challenges us to adjust the solution format to consider the multiplicity of the roots, resulting in:

• \( y_h(x) = C_1 e^{-2x} + C_2 x e^{-2x} \)

Thus, identifying these roots and incorporating them correctly is vital for solving the differential equation.