Question

In Exercises 20-23 solve the initial value problem and plot the solution. $$ y^{\prime \prime}+3 y^{\prime}-4 y=e^{2 x}(7+6 x), \quad y(0)=2, \quad y^{\prime}(0)=8 $$

Short answer

Question: Determine the solution to the initial value problem $$y'' + 3y' - 4y = e^{2x}(7+6x),$$ with initial conditions $$y(0) = 2$$ and $$y'(0) = 8$$, and plot the solution. Answer: The solution to the initial value problem is given by $$y(x) = -e^{-4x} + 2e^{x} + (2x^{2} + 2x+1) e^{2x}$$. Use graphing tools or software like desmos, GeoGebra, or MATLAB to plot the solution for this differential equation.

Step-by-step solution

Identify the Complementary Function

To find the complementary function, we need to solve the homogeneous equation: $$ y'' + 3y' - 4y = 0 $$ The characteristic equation for this second-order linear homogeneous differential equation is: $$ r^2 + 3r - 4 = 0 $$ Solve for 'r' to find the complementary function.

Solve the Characteristic Equation

The characteristic equation we derived in step 1 is a quadratic equation: $$ r^2 + 3r - 4 = 0 $$ We can factor this equation to find the roots: $$ (r + 4)(r - 1) = 0 $$ So, the roots are r1 = -4 and r2 = 1.

Write the Complementary Function

Using the roots found in step 2, we can write the complementary function as a linear combination of exponentials: $$ y_c(x) = C_1 e^{-4x} + C_2 e^{x} $$ Where 'C1' and 'C2' are constants to be determined using the initial conditions.

Find the Particular Solution

To find the particular solution, we use a guess method. Since the non-homogeneous term is of the form \(e^{2x}(7+6x)\), we will guess a solution of the form: $$ y_p(x) = (Ax^{2} + Bx+C) e^{2x} $$ Compute the first and second derivatives of the particular solution and then substitute the particular solution and its derivatives into the given initial value problem to solve for the coefficients 'A', 'B', and 'C'.

Substitution and Solving for Coefficients

Find the first derivative: $$ y_p'(x) = 2(Ax^2 + Bx + C)e^{2x} + (2Ax+B)e^{2x} $$ Now, find the second derivative: $$ y_p''(x) = (2Ax+2B)e^{2x} + 4(Ax^{2} + Bx + C)e^{2x} + 4(2Ax + B)e^{2x} $$ Now, substitute the particular solution and its derivatives into the given initial value problem: $$ y_p''(x) + 3y_p'(x) - 4y_p(x) = e^{2 x}(7+6 x) $$ Upon substitution, we obtain: $$ (Ax^2 + Bx + C)e^{2x} = (7+6x)e^{2x} $$ After that, we solve for coefficients (A, B, C) which gives: A = 2, B = 2, C = 1 The particular solution is: $$ y_p(x) = (2x^{2} + 2x+1) e^{2x} $$

General Solution

Now that we have both the complementary function and the particular solution, we can write the general solution as the sum of the complementary and particular solutions: $$ y(x) = y_c(x) + y_p(x) = C_1 e^{-4x} + C_2 e^{x} + (2x^{2} + 2x+1) e^{2x} $$

Apply Initial Conditions

Now we will apply the initial conditions to find the constants 'C1' and 'C2': For y(0) = 2: $$ 2 = C_1 e^{0} + C_2 e^{0} + (0^2 + 0 + 1)e^{0} $$ Simplifying, we get: $$ C_1 + C_2 = 1 $$ For y'(0) = 8: $$ 8 = -4C_1 e^{0} + C_2 e^{0} + (4 + 2)e^{0} $$ Simplifying, we obtain: $$ -4C_1 + C_2 = 2 $$ Now, we solve the system of equations: $$ C_1 + C_2 = 1 \text{ and } -4C_1 + C_2 = 2 $$ This gives us C1 = -1 and C2 = 2.

Final Solution

Now that we have found the constants 'C1' and 'C2', we can write the final solution: $$ y(x) = -e^{-4x} + 2e^{x} + (2x^{2} + 2x+1) e^{2x} $$ This is the solution to the initial value problem. To plot the solution for this differential equation, you can use various graphing tools or software like desmos, GeoGebra, or MATLAB.

Key concepts

Complementary Function

The complementary function is an essential part of solving a non-homogeneous differential equation. It represents the general solution of the associated homogeneous differential equation. In this problem, the homogeneous equation is given by:
\[ y'' + 3y' - 4y = 0 \]
To find the complementary function, you first solve this equation by finding its roots. This is done by deriving its characteristic equation, which in this case is a quadratic equation:

• \( r^2 + 3r - 4 = 0 \)

After solving for the roots of the characteristic equation, we have \( r_1 = -4 \) and \( r_2 = 1 \). Using these roots, the complementary function is expressed as:
\[ y_c(x) = C_1 e^{-4x} + C_2 e^{x} \]
where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. This forms the foundation of the solution by addressing the homogeneous aspect of the problem.

Particular Solution

Once you determine the complementary function, the next step is to find a particular solution for the non-homogeneous differential equation. The non-homogeneous part of the equation is identified as \( e^{2x}(7+6x) \). For such terms, the method of undetermined coefficients is often used.
In this scenario, you assume a solution form that mimics the non-homogeneous part:
\[ y_p(x) = (Ax^2 + Bx + C) e^{2x} \]
To find the right coefficients \( A, B, \) and \( C \), substitute \( y_p(x) \) into the differential equation and let it equal the non-homogeneous part. Solving this will give:

• A = 2
• B = 2
• C = 1

Thus, the particular solution becomes:
\[ y_p(x) = (2x^2 + 2x + 1) e^{2x} \]
Ultimately, the particular solution fills in the gaps left by the complementary function for the solution to meet the specific non-homogeneous criteria.

Characteristic Equation

The characteristic equation is critical in the process of solving homogeneous differential equations. This equation arises from the coefficients of the derivatives in the homogeneous differential equation.
In the example, starting from the homogeneous equation:
\[ y'' + 3y' - 4y = 0 \]
We derive the characteristic equation by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. This transformation gives:
\[ r^2 + 3r - 4 = 0 \]
This is a quadratic equation, and its roots can be found through factoring, completing the square, or using the quadratic formula. Here, it factors nicely to:

• (r + 4)(r - 1) = 0

This gives the roots \( r_1 = -4 \) and \( r_2 = 1 \), which represent different exponential terms in the complementary function. Each root corresponds to a term in the form \( e^{rx} \), crucial for solving differential equations.

Homogeneous Differential Equation

A homogeneous differential equation is an equation in which every term is a function of the dependent variable and its derivatives. These types of equations imply that no term in the equation is independent or a constant.
The equation \( y'' + 3y' - 4y = 0 \) is an example of a homogeneous differential equation. It sets the foundation for finding the complementary function, which is purely based on these homogeneous equations.

• All terms involve the function \( y \) or its derivatives.
• The right-hand side equals zero.

The primary goal of solving this equation is to obtain the complementary function which reflects all possible solutions without external influence. By determining the complementary function, you are solving the 'natural' aspect of the system described by the differential equation.
Ultimately, understanding the homogeneous differential equation is vital as it provides the initial structure needed before incorporating any particular solutions to form the general solution.