Question

Solve the initial value problem and graph the solution. $$ y^{\prime \prime}+8 y^{\prime}+7 y=10 e^{-2 x}, \quad y(0)=-2, y^{\prime}(0)=10 $$

Short answer

Question: Solve the initial value problem for the given second-order, linear, non-homogeneous differential equation with constant coefficients and initial conditions: $$ y^{\prime \prime} + 8y^{\prime} + 7y = 10e^{-2x}, \; y(0) = -2, \; y'(0) = 10 $$ Answer: The exact solution for the initial value problem is: $$ y(x) = \frac{1}{2} e^{-x} - \frac{5}{2} e^{-7x} - \frac{5}{2}xe^{-2x} $$

Step-by-step solution

Find the complementary function

To find the complementary function, we will solve the homogeneous equation. $$ y^{\prime \prime} + 8y^{\prime} + 7y = 0 $$ This is a second-order linear differential equation with constant coefficients, so we will solve the characteristic equation. $$ m^2 + 8m + 7 = 0 $$ Solving for m, we have: $$ (m + 1)(m + 7) = 0 $$ So, the complementary function is given by: $$ y_c(x) = C_1 e^{-x} + C_2 e^{-7x} $$

Find a particular solution

Given the form of the non-homogeneous term, \(10e^{-2x}\), we make an educated guess and propose the following form for a particular solution: $$ y_p(x) = A x e^{-2x} $$ Differentiating \(y_p(x)\) with respect to \(x\), we get: $$ y_p^{'}(x) = A e^{-2x} - 2Axe^{-2x} $$ Differentiating the first derivative with respect to \(x\), we obtain the second derivative: $$ y_p^{''}(x) = -2Ae^{-2x} + 4Axe^{-2x} $$ Substituting \(y_p(x)\), \(y_p^{'}(x)\), and \(y_p^{''}(x)\) into the non-homogeneous equation, we get: $$ (-2Ae^{-2x} + 4Axe^{-2x}) + 8(A e^{-2x} - 2Axe^{-2x}) + 7(A xe^{-2x}) = 10 e^{-2x} $$ Simplifying, we have: $$ (4A - 14A)xe^{-2x} = 10 e^{-2x} $$ Comparing the coefficients, we find that \(A = -\frac{5}{2}\). Therefore, the particular solution is: $$ y_p(x) = -\frac{5}{2}xe^{-2x} $$

Determine the general solution

To find the general solution to the initial value problem, we combine the complementary function and the particular solution: $$ y(x) = y_c(x) + y_p(x) = C_1 e^{-x} + C_2 e^{-7x} - \frac{5}{2}xe^{-2x} $$

Apply the initial conditions

To obtain the exact solution for this initial value problem, we apply the initial conditions \(y(0) = -2\) and \(y'(0) = 10\). For the first initial condition, we have: $$ y(0) = C_1 e^{0} + C_2 e^{0} - 0 = -2 $$ This gives us \(C_1 + C_2 = -2\). For the second initial condition, we first need the derivative of the general solution: $$ y^{'}(x) = -C_1 e^{-x} - 7C_2 e^{-7x} + \frac{5}{2} e^{-2x} - 5xe^{-2x} $$ Now we apply the initial condition \(y'(0) = 10\): $$ y^{'}(0) = -C_1 - 7C_2 + \frac{5}{2} = 10 $$ Solve the system of equations: $ \begin{cases} C_1 + C_2=-2 \\ - C_1 -7 C_2=-\frac{15}{2} \end{cases} $ We get \(C_1 = \frac{1}{2}\) and \(C_2 = -\frac{5}{2}\). So, the exact solution for the initial value problem is: $$ y(x) = \frac{1}{2} e^{-x} - \frac{5}{2} e^{-7x} - \frac{5}{2}xe^{-2x} $$ To graph the solution, plot \(y(x) = \frac{1}{2} e^{-x} - \frac{5}{2} e^{-7x} - \frac{5}{2}xe^{-2x}\) on a graphing calculator or suitable software.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation along with specified values at a given point. These specified values are known as initial conditions. In this exercise, the differential equation is a second-order linear equation:

• \( y'' + 8y' + 7y = 10e^{-2x} \)
• with the initial conditions \( y(0) = -2 \) and \( y'(0) = 10 \).

The goal is to find a solution that fits the differential equation and meets the initial conditions provided. Solving an IVP often involves finding a general solution to the differential equation and then applying the initial conditions to find specific values for any constants in the solution. Once determined, this solution will satisfy both the differential equation and the initial conditions.

Complementary Function

The complementary function is a solution to the homogeneous version of a differential equation. For the equation in this exercise:

• The homogeneous version is \( y'' + 8y' + 7y = 0 \).

To find the complementary function, we typically solve the characteristic equation. This involves:

• Forming the characteristic equation from the homogeneous differential equation: \( m^2 + 8m + 7 = 0 \).
• Solving for the roots, which in this case are \( m = -1 \) and \( m = -7 \).

These roots help us build the complementary function, \( y_c(x) = C_1 e^{-x} + C_2 e^{-7x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants. This function forms the backbone of the general solution.

Particular Solution

A particular solution is a specific solution to a non-homogeneous differential equation that accounts for the non-zero part of the equation. In this exercise, the non-homogeneous term is \( 10e^{-2x} \).
We guess a form for the particular solution based on the non-homogeneous term's structure; here, we propose \( y_p(x) = Ax e^{-2x} \).
By substituting this guess into the differential equation and simplifying, we solve for the constant \( A \). Comparing coefficients gives us \( A = -\frac{5}{2} \), resulting in the particular solution \( y_p(x) = -\frac{5}{2}xe^{-2x} \).
This part of the solution addresses the influence of the non-homogeneous term in the differential equation.

Characteristic Equation

The characteristic equation is crucial for solving linear differential equations with constant coefficients. For the homogeneous equation \( y'' + 8y' + 7y = 0 \), the characteristic equation is:

• \( m^2 + 8m + 7 = 0 \)

This is a quadratic equation, which we solve for its roots. These roots dictate the form of the complementary function.
In this example:

• The characteristic roots are \( m = -1 \) and \( m = -7 \).

The roots define the exponential terms \( e^{-x} \) and \( e^{-7x} \) in the complementary function. The ability to find and correctly use the characteristic equation roots is critical in solving differential equations effectively.