Problem 72 - Calculus 3

Question

Solve the differential equation using either the method of undetermined coefficients or the variation of parameters. $$y^{\prime \prime}+3 y^{\prime}-4 y=2 e^{x}$$

Step-by-Step Solution

Answer

The solution is \( y = C_1 e^{-4x} + C_2 e^{x} + \frac{2}{5} x e^{x} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is a linear non-homogeneous second-order differential equation of the form \( y'' + 3y' - 4y = 2e^x \). We will solve it using the method of undetermined coefficients.

Step 2: Solve the Homogeneous Equation

Solve the corresponding homogeneous equation \( y'' + 3y' - 4y = 0 \). First, find the characteristic equation: \( r^2 + 3r - 4 = 0 \).Factor the quadratic equation: \((r + 4)(r - 1) = 0 \).The roots are \( r_1 = -4 \) and \( r_2 = 1 \). Therefore, the complementary solution is \( y_c = C_1 e^{-4x} + C_2 e^{x} \).

Step 3: Guess a Particular Solution

Since the non-homogeneous term is \( 2e^x \), we guess a particular solution of the form \( y_p = Ax e^{x} \). We use \( xe^x \) instead of \( e^x \) because \( e^x \) is a solution to the homogeneous equation.

Step 4: Differentiate the Particular Solution

Compute the first and second derivatives of \( y_p = Ax e^{x} \):\( y_p' = A e^{x} + A x e^{x} \)\( y_p'' = 2A e^{x} + A x e^{x} \).

Step 5: Substitute and Solve for A

Substitute \( y_p, y_p', y_p'' \) into the original differential equation:\( (2A e^{x} + A x e^{x}) + 3(A e^{x} + A x e^{x}) - 4(A x e^{x}) = 2 e^{x} \).Simplify and combine like terms:\( 5A e^{x} = 2 e^{x} \).Solve for \( A \):\( 5A = 2 \) which gives \( A = \frac{2}{5} \).

Step 6: Write General Solution

Combine the complementary solution and the particular solution:\( y = y_c + y_p = C_1 e^{-4x} + C_2 e^{x} + \frac{2}{5} x e^{x} \).

Key Concepts

Method of Undetermined Coefficients

The method of undetermined coefficients is a popular technique used for solving linear non-homogeneous differential equations. It involves assuming a form for the particular solution based on the non-homogeneous term, and then determining the coefficients by substituting back into the original equation. This method is best suited for equations where the non-homogeneous part is a simple function like a polynomial, exponential, sine, or cosine function.

To use this method, follow these steps:

• Identify the form of the non-homogeneous term. For instance, in our given differential equation, the term is an exponential, specifically, 2e^x.
• Guess a form for the particular solution that matches the structure of the non-homogeneous term. However, make sure it does not match any part of the complementary solution derived from the homogeneous equation.
• Determine coefficients by substituting the guessed particular solution into the differential equation and equating coefficients of similar terms on both sides.

This technique is efficient and avoids complex integrations, making it a convenient choice for the specific type of differential equation in our example.

Homogeneous Equations

Homogeneous differential equations play a crucial role in the overall solution of non-homogeneous systems. A homogeneous equation is derived from the original differential equation by setting the non-homogeneous term to zero. For example, from the original equation \( y'' + 3y' - 4y = 2e^x \), we form the homogeneous equation \( y'' + 3y' - 4y = 0 \).

Solving a homogeneous equation involves:

• Finding the characteristic equation, which is derived from the associated polynomial, here being \( r^2 + 3r - 4 = 0 \).
• Solving the characteristic equation to find its roots. In this case, the roots are \( r_1 = -4 \) and \( r_2 = 1 \).
• Writing the complementary (general) solution using these roots. The complementary solution is \( y_c = C_1 e^{-4x} + C_2 e^{x} \).

This solution represents all possible solutions if the system were entirely homogeneous, providing the foundational component needed to construct the general solution of the original differential equation.

Particular Solution

The particular solution is crucial for equating a non-homogeneous differential equation. It directly accounts for the part of the equation caused by external forces or inputs, represented by the non-homogeneous term. In our original differential equation, the particular solution deals with the right-hand side term, 2e^x.

When the guessed particular solution is similar to any part of the complementary solution, it must be modified to ensure linear independence. In our case, since \( e^x \) is a solution to the homogeneous equation, the assumption of the particular solution is altered to \( A x e^{x} \) instead of just \( A e^{x} \).

This adjustment ensures that the final solution does not simply blend back into the homogeneous solution, nor does it duplicate its components, thereby maintaining the integrity of the solution set.

Characteristic Equation

The characteristic equation is foundational in solving linear differential equations, particularly when separating the effects of the homogeneous part of the equation from the non-homogeneous part. It is formed by temporarily setting the non-homogeneous terms to zero and focusing purely on the coefficients of the derivatives.

In the given case, the characteristic equation is obtained by replacing \( y'' \), \( y' \), and \( y \) with powers of \( r \), which is an algebraic strategy. Here, \( y'' + 3y' - 4y = 0 \) transforms into \( r^2 + 3r - 4 = 0 \).

Solve this quadratic equation using:

• Factoring: \((r + 4)(r - 1) = 0\).
• Identifying roots: \( r_1 = -4 \) and \( r_2 = 1 \).

These roots allow us to express the complementary solution, offering insights into how the system's intrinsic characteristics behave over time when unperturbed by external inputs. The characteristic equation thus guides the form of the general solution.