Problem 31 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$y^{\prime \prime}+4 y^{\prime}+5 y=35 e^{-4 x}, \quad y(0)=-3, y^{\prime}(0)=1$$

Step-by-Step Solution

Answer

The solution is \( y(x) = e^{-2x}(-10\cos(x) - 5\sin(x)) + 7e^{-4x} \).

Step 1: Identifying the Type of Differential Equation

The given equation is a second-order linear non-homogeneous differential equation with constant coefficients, denoted as:\[ y'' + 4y' + 5y = 35e^{-4x} \]

Step 2: Solving the Homogeneous Equation

First, solve the associated homogeneous equation:\[ y'' + 4y' + 5y = 0 \]Assume a solution of the form \( y_h = e^{rx} \). The characteristic equation is:\[ r^2 + 4r + 5 = 0 \]Solving using the quadratic formula gives:\[ r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i \]This corresponds to a solution:\[ y_h = e^{-2x}(C_1\cos(x) + C_2\sin(x)) \]

Step 3: Finding a Particular Solution

To find a particular solution, \( y_p \), propose a form that resembles the non-homogeneous part, i.e., \( y_p = Ae^{-4x} \). Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original equation:\[ y_p' = -4Ae^{-4x} , \quad y_p'' = 16Ae^{-4x} \]\[ 16Ae^{-4x} - 16Ae^{-4x} + 5Ae^{-4x} = 35e^{-4x} \]Simplifying gives:\[ 5A = 35 \rightarrow A = 7 \]Thus, \( y_p = 7e^{-4x} \).

Step 4: Constructing the General Solution

The general solution is the sum of the homogeneous and particular solutions:\[ y(x) = y_h + y_p = e^{-2x}(C_1\cos(x) + C_2\sin(x)) + 7e^{-4x} \]

Step 5: Applying Initial Conditions

Use the initial conditions to solve for constants \( C_1 \) and \( C_2 \). Use \( y(0) = -3 \):\[ e^{0}(C_1\cos(0) + C_2\sin(0)) + 7e^{0} = -3 \]\[ C_1 + 7 = -3 \quad\Rightarrow\quad C_1 = -10 \]Now, use \( y'(0) = 1 \):The derivative is:\[ y' = e^{-2x}(-2C_1\cos(x) - C_1\sin(x) - 2C_2\sin(x) + C_2\cos(x)) - 14e^{-4x} \]\[ y'(0) = 1 = -2(-10) \cdot 1 + C_2 \cdot 1 - 14 \cdot 1 \]\[ 1 = 20 + C_2 - 14 \quad\Rightarrow\quad C_2 = -5 \]

Step 6: Final Solution

Substitute back \( C_1 = -10 \) and \( C_2 = -5 \) into the general solution:\[ y(x) = e^{-2x}(-10\cos(x) - 5\sin(x)) + 7e^{-4x} \]

Key Concepts

Understanding Initial-Value Problems

Initial-value problems involve finding a specific solution to a differential equation that also satisfies given conditions, often at a particular point. This means you're not just solving the equation in general, but you're also making sure it fits certain criteria from the start.

• For our exercise, the initial conditions are \( y(0) = -3 \) and \( y'(0) = 1 \). These tell us the value of the function and its derivative at \( x = 0 \).
• These conditions help us find the exact values for the constants in the solution of the differential equation. Without them, you'd end up with a family of solutions rather than one specific answer.

By applying the initial conditions, we can narrow down the infinite possibilities to the single solution that fits our problem's criteria.

Exploring Second-Order Linear Differential Equations

Second-order linear differential equations involve terms with second derivatives and can usually be written in this standard form: \( ay'' + by' + cy = g(x) \). The one in our problem is:

• \( y'' + 4y' + 5y = 35e^{-4x} \).
• The coefficients (like 4 and 5) are constants here, which simplifies many things.

Such equations are widespread in physics and engineering, modeling situations like motion and wave propagation.Second-order means the highest derivative is the second derivative. The 'linear' part describes the equation's structure, where all terms involving the function and its derivatives are to the first power.To solve these equations, you must find the homogeneous solution \( y_h \) and a particular solution \( y_p \). These are combined to form the general solution.

Non-Homogeneous Differential Equations Demystified

Non-homogeneous differential equations have a term that isn’t dependent on the solution or its derivatives. This term is known as the non-homogeneous part, like \( 35e^{-4x} \) in our example.

• To tackle these, you find a homogeneous solution \( y_h \) first by solving the related homogeneous equation (omit the non-homogeneous part).
• Next, propose a particular solution \( y_p \) by mimicking the form of the non-homogeneous term.

The art comes in choosing \( y_p \). If your choice is correct, substituting back into the differential equation will simplify nicely, letting you solve for any constants in \( y_p \).Finally, you combine \( y_h \) and \( y_p \) to form the general solution. This general solution is the backbone for applying initial conditions.

Decoding the Characteristic Equation

The characteristic equation is a crucial part of solving the homogeneous portion of a differential equation. For our equation, you assume a solution of the form \( y_h = e^{rx} \).

• This transforms the differential equation into a polynomial equation in terms of \( r \), known as the characteristic equation.
• For our problem, it is \( r^2 + 4r + 5 = 0 \).

Solving the characteristic equation usually involves algebra, such as factoring or using the quadratic formula, as seen here:\[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i \].The solutions to this equation, \( -2 \pm i \), indicate complex roots, leading to a particular type of homogeneous solution:\( y_h = e^{-2x} (C_1 \cos(x) + C_2 \sin(x)) \).This process reveals the behavior of the system described by the differential equation, such as oscillations or exponential growth/decay.