Problem 81 - Calculus Volume 3

Question

Find the unique solution satisfying the differential equation and the initial conditions given, where \(y_{p}(x)\) is the particular solution. \( y^{\prime \prime}-7 y^{\prime}=4 x e^{7 x}, \quad y_{p}(x)=\frac{2}{7} x^{2} e^{7 x}-\frac{4}{49} x e^{7 x}\) \(y(0)=-1, \quad y^{\prime}(0)=0\)

Step-by-Step Solution

Answer

The unique solution is \( y(x) = -1 + \frac{2}{7}x^2 e^{7x} - \frac{4}{49}x e^{7x} \).

Step 1: Determine the homogeneous solution

The characteristic equation for the homogeneous equation \( y'' - 7y' = 0 \) is \( r^2 - 7r = 0 \). This can be factored to \( r(r - 7) = 0 \), giving roots \( r_1 = 0 \) and \( r_2 = 7 \). Thus, the complementary solution is \( y_h(x) = C_1 + C_2 e^{7x} \).

Step 2: General solution for the differential equation

The general solution \( y(x) \) is the sum of the complementary (homogeneous) solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) = C_1 + C_2 e^{7x} + \frac{2}{7}x^2 e^{7x} - \frac{4}{49} x e^{7x}. \]

Step 3: Apply initial condition: \( y(0) = -1 \)

Substitute \( x = 0 \) into the general solution:\[ y(0) = C_1 + C_2 = -1. \]From the particular solution terms, both vanish when \( x = 0 \), hence \( C_1 + C_2 = -1. \)

Step 4: Apply initial condition: \( y' (0) = 0 \)

First, find the derivative \( y'(x) \):\[ y'(x) = C_2 \, 7 e^{7x} + \left(\frac{2}{7}(2x)e^{7x} + \frac{2}{7}x^2 (7e^{7x}) - \frac{4}{49} e^{7x} - \frac{4}{49}x(7e^{7x})\right). \]At \( x = 0 \), this simplifies to \( y'(0) = 7C_2 + 0 = 0 \).Therefore, \( C_2 = 0 \).

Step 5: Solve for \( C_1 \)

Using \( C_2 = 0 \) in the equation from Step 3, \( C_1 + 0 = -1 \), we find that \( C_1 = -1 \).

Step 6: Write the unique solution

Now that \( C_1 \) and \( C_2 \) are known, substitute them into the general solution:\[ y(x) = -1 + \frac{2}{7}x^2 e^{7x} - \frac{4}{49} x e^{7x}. \]

Key Concepts

Initial Conditions

Initial conditions are values that help us determine the specific solution to a differential equation. These are given values for the function and its derivatives at a particular point—usually when the independent variable (like time or position) equals zero.
For example, the initial conditions here are:

• \( y(0) = -1 \), which is the value of the function when \( x = 0 \).
• \( y'(0) = 0 \), indicating the initial rate of change of the function is zero at \( x = 0 \).

By applying these initial conditions, we can solve for the constants in the general solution, finding the unique solution that satisfies the original differential equation.

Homogeneous Solution

In the context of differential equations, the homogeneous solution refers to the solution of the homogeneous part of the differential equation. This means ignoring the terms that are not multiplied by the function or its derivatives.
In our exercise, the homogeneous differential equation is:\[y^{\prime\prime} - 7 y^{\prime} = 0. \]This leads us to solve the characteristic equation, which we form by replacing the derivatives with powers of \( r \). Solving \( r^2 - 7r = 0 \) yields roots, which help us construct the homogeneous solution:

• \( r_1 = 0 \), yielding the term \( C_1 \)
• \( r_2 = 7 \), yielding \( C_2 e^{7x} \)

Hence, the homogeneous solution is \( y_h(x) = C_1 + C_2 e^{7x} \). This part of the solution captures the nature of the system without any external forces acting on it.

Particular Solution

A particular solution is a solution to the non-homogeneous differential equation that specifically accounts for the non-homogeneous (or external) part of the equation.
In our problem, the external part is \( 4x e^{7x} \). We find a particular solution that includes this term explicitly.
The given particular solution is:\[y_p(x) = \frac{2}{7}x^2 e^{7x} - \frac{4}{49} x e^{7x}.\]This solution is tailored to neutralize the non-homogeneous part when substituted back into the original equation. Once found, the particular solution is combined with the homogeneous solution to form the general solution.

Characteristic Equation

The characteristic equation is crucial in finding the homogeneous solution of a linear differential equation with constant coefficients. It arises from substituting trial solutions of the form \( y = e^{rx} \) into the homogeneous equation.
For the differential equation:\[y^{\prime\prime} - 7 y^{\prime} = 0,\]the characteristic equation is set up by treating \( y'' \) as \( r^2 \) and \( y' \) as \( r \): \[r^2 - 7r = 0.\]We solve this polynomial equation by factoring it to get:

• \( r(r - 7) = 0 \)

The roots \( r_1 = 0 \) and \( r_2 = 7 \) provide the basis for the solutions of the homogeneous part of the equation, which together form the general expression for \( y_h(x) \). This method allows us to systematically find the essential components needed to solve the complete differential equation.