Problem 80 - Calculus 3

Question

Find the unique solution satisfying the differential equation and the initial conditions given, where \(y_{p}(x)\) is the particular solution. \(y^{\prime \prime}-2 y^{\prime}+y=12 e^{x}\), \(\mathrm{y}_{p}(x)=6 x^{2} e^{x}\), \(y(0)=-1, y^{\prime}(0)=0\)

Step-by-Step Solution

Answer

The unique solution is \( y(x) = (6x^2 + x - 1)e^x \).

Step 1: Homogeneous Solution

First, solve the homogeneous differential equation associated with the given equation: \( y'' - 2y' + y = 0 \). The characteristic equation is \( r^2 - 2r + 1 = 0 \), which factors as \( (r - 1)^2 = 0 \). This gives a repeated root \( r = 1 \). Therefore, the complementary (homogeneous) solution is \( y_h(x) = (C_1 + C_2x)e^x \), where \( C_1 \) and \( C_2 \) are constants.

Step 2: Particular Solution

We are given the particular solution \( y_p(x) = 6x^2 e^x \). Therefore, the general solution combining both homogeneous and particular parts is \( y(x) = y_h(x) + y_p(x) = (C_1 + C_2x)e^x + 6x^2 e^x \).

Step 3: Apply Initial Conditions

Use the initial conditions \( y(0) = -1 \) and \( y'(0) = 0 \). First, calculate \( y(0) \):\ \((C_1 + C_2(0))e^0 + 6(0)^2 e^0 = C_1 = -1\). Next, calculate the first derivative \( y'(x) \) and use \( y'(0) = 0 \). Start by differentiating: \( y'(x) = (C_2x + C_1)e^x + (C_1 + C_2x)e^x + 12xe^x + 6x^2e^x \). Evaluate at \( x = 0 \): \((C_2(0) + C_1)e^0 + (C_2 e^0 + 12(0)e^0 + 6(0)^2 e^0 ) = C_1 + C_2 = 0\). Substituting \( C_1 = -1 \), we find \( C_2 = 1 \).

Step 4: Write the Final Solution

With \( C_1 = -1 \) and \( C_2 = 1 \), the solution satisfying the initial conditions is \( y(x) = ((-1) + x)e^x + 6x^2 e^x = (x - 1)e^x + 6x^2 e^x \). Simplify to \( y(x) = (x - 1 + 6x^2)e^x = (6x^2 + x - 1)e^x \).

Key Concepts

Homogeneous Solution

In differential equations, the homogeneous solution refers to solving the associated homogeneous equation. For the equation given, we dealt with the homogeneous part, where no external functions, like exponential terms, are added. Our task was to solve the equation: \( y'' - 2y' + y = 0 \).To find the homogeneous solution, we first tackle the characteristic equation. This is obtained by replacing derivatives with powers of \( r \).

• The equation becomes: \( r^2 - 2r + 1 = 0 \).
• On factoring, we have \( (r - 1)^2 = 0 \).
• This gives a repeated root: \( r = 1 \).

The repeated root results in a homogeneous solution that takes the form:\( y_h(x) = (C_1 + C_2x)e^x \),where \( C_1 \) and \( C_2 \) are constants. This solution captures all solutions considering only the homogeneous part of the differential equation.

Particular Solution

The particular solution, denoted as \( y_p(x) \), is the part of the solution that accounts for any non-homogeneous aspects of the differential equation. In our exercise, the function \( 12e^x \) is the external (non-homogeneous) part.
For this equation, we are provided with a particular solution:\( y_p(x) = 6x^2 e^x \).The particular solution represents a specific solution to the entire non-homogeneous equation, \( y'' - 2y' + y = 12e^x \).
When determining a particular solution:

• Methods such as undetermined coefficients or variation of parameters can be used.
• Here, the particular solution is given directly.

By combining the homogeneous and particular solutions, we form the general solution to the differential equation, which provides a more complete picture.

Characteristic Equation

The characteristic equation is crucial when solving linear differential equations, especially when dealing with constant coefficients. It helps us find the solutions for the homogeneous equation by turning the differential equation into an algebraic one.
Here's how it's derived for this specific problem:

• Original differential equation: \( y'' - 2y' + y = 0 \).
• Assuming a solution of the form \( y = e^{rx} \), we substitute to find the characteristic equation.
• This leads to: \( r^2 - 2r + 1 = 0 \).
• Factoring gives \( (r-1)^2 = 0 \), indicating a repeated root \( r = 1 \).

This process simplifies the determination of the homogeneous solution by providing the necessary roots for constructing the solution.

Initial Conditions

Initial conditions are used to determine the specific solution of a differential equation that fits given criteria. They provide values at a specific point, usually \( x = 0 \), to find the unknown constants in the general solution.
For our example, we're given:

• \( y(0) = -1 \)
• \( y'(0) = 0 \)

Using these conditions:1. We substitute into the general solution to find \( C_1 \) and \( C_2 \). - Setting \( y(0) = -1 \) results in \( C_1 = -1 \).2. Differentiating to find \( y'(x) \), we use \( y'(0) = 0 \).
- This gives \( C_2 = 1 \), since \( C_1 + C_2 = 0 \).By setting our constants, we derive the unique solution that fits the differential equation and satisfies the initial conditions, ensuring that this solution behaves as expected at \( x = 0 \).