Problem 59 - Calculus 3

Question

Solve the following equations using the method of undetermined coefficients. $$y^{\prime \prime}-4 y^{\prime}+4 y=8 x^{2}+4 x$$

Step-by-Step Solution

Answer

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

Step 1: Write the Characteristic Equation

To solve the homogeneous equation associated with the differential equation, start by writing the characteristic equation.The homogeneous equation is: \[ y'' - 4y' + 4y = 0 \]The characteristic equation is: \[ r^2 - 4r + 4 = 0 \]

Step 2: Solve the Characteristic Equation

Solve the characteristic equation \( r^2 - 4r + 4 = 0 \) using the quadratic formula or factoring.The equation can be factored as:\[ (r-2)^2 = 0 \]Thus, we find a repeated root: \( r = 2 \).

Step 3: Write the Complementary Solution

Based on the repeated root, the complementary solution is:\[ y_c = (C_1 + C_2 x) e^{2x} \]Here, \( C_1 \) and \( C_2 \) are constants to be determined by initial conditions if given.

Step 4: Assume a Particular Solution

For the non-homogeneous term \( 8x^2 + 4x \), assume a particular solution of the form:\[ y_p = Ax^2 + Bx + C \]

Step 5: Differentiate the Particular Solution

Differentiate \( y_p = Ax^2 + Bx + C \) with respect to \( x \).First derivative:\[ y_p' = 2Ax + B \]Second derivative:\[ y_p'' = 2A \]

Step 6: Substitute and Simplify

Substitute \( y_p, y_p', y_p'' \) into the original differential equation:\[ 2A - 4(2Ax + B) + 4(Ax^2 + Bx + C) = 8x^2 + 4x \]Simplify the expression:\[ 2A - 8Ax - 4B + 4Ax^2 + 4Bx + 4C = 8x^2 + 4x \]

Step 7: Equate Coefficients

Equate the coefficients of \( x^2, x, \) and the constant terms from both sides of the equation:For \( x^2: \) \( 4A = 8 \) For \( x: \) \( -8A + 4B = 4 \) For the constant term: \( 2A - 4B + 4C = 0 \)Solve these equations for \( A, B, \) and \( C \).

Step 8: Solve for Coefficients

First, solve \( 4A = 8 \) to find \( A = 2 \).Next, substitute \( A = 2 \) into the equation \( -8A + 4B = 4 \):\[ -8(2) + 4B = 4 \]\[ -16 + 4B = 4 \]\[ 4B = 20 \]\[ B = 5 \]Finally, substitute \( A = 2 \) and \( B = 5 \) into the equation \( 2A - 4B + 4C = 0 \):\[ 2(2) - 4(5) + 4C = 0 \]\[ 4 - 20 + 4C = 0 \]\[ 4C = 16 \]\[ C = 4 \]

Step 9: Write the Particular Solution

The particular solution based on the values of \( A, B, \) and \( C \) is:\[ y_p = 2x^2 + 5x + 4 \]

Step 10: Form the General Solution

The general solution is the sum of the complementary and particular solutions:\[ y = y_c + y_p = (C_1 + C_2 x) e^{2x} + 2x^2 + 5x + 4 \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations express a relationship between a function and its rates of change, and they are pivotal in understanding various phenomena in fields like physics, biology, and engineering. In simple terms, they help us understand how a quantity changes concerning another.

In the given problem, we are dealing with a second-order linear differential equation, where the highest derivative is the second derivative of the function, represented as \( y'' \). The equation is:

• \( y'' - 4y' + 4y = 8x^2 + 4x \)

This equation is non-homogeneous because it includes functions of \( x \) on the right side. Solving such equations typically involves finding a complementary solution for the associated homogeneous equation and a particular solution for the non-homogeneous part.

Complementary Solution

The complementary solution is a part of the overall solution to a differential equation and solves the corresponding homogeneous equation. For our example, the homogeneous part is:

• \( y'' - 4y' + 4y = 0 \)

Finding the complementary solution involves solving the characteristic equation, which is derived by replacing the derivatives in the homogeneous equation with powers of \( r \). This results in a polynomial equation:

• \( r^2 - 4r + 4 = 0 \)

Factoring or using the quadratic formula, we find the characteristic roots. For our equation, the root is \( r = 2 \) (a repeated root).
The complementary solution is then:

• \( y_c = (C_1 + C_2 x) e^{2x} \)

where \( C_1 \) and \( C_2 \) are arbitrary constants. This form accommodates the repeated root by including a linear term, which is necessary to account for the multiplicity of the root.

Particular Solution

The particular solution addresses the non-homogeneous part of a differential equation, providing a specific solution based on the form of the non-homogeneous term. For the given problem, we have the term \( 8x^2 + 4x \).

One efficient method to assume the form of a particular solution is using the method of undetermined coefficients. This involves assuming a potential solution form that reflects the non-homogeneous component. Here, we propose:

• \( y_p = Ax^2 + Bx + C \)

To find the constants \( A, B, \) and \( C \), substitute \( y_p \), and its derivatives back into the original differential equation and match coefficients with the like terms on the right side.

Once calculated, the particular solution turns out to be:

• \( y_p = 2x^2 + 5x + 4 \)

Together with the complementary solution, this helps us construct the complete solution to the differential equation.

Characteristic Equation

The characteristic equation is derived from the homogeneous part of a differential equation and is crucial for finding the complementary solution. It is formed by replacing the derivatives with powers of an auxiliary variable, generally denoted as \( r \).

For the differential equation given, the homogeneous equation is:

• \( y'' - 4y' + 4y = 0 \)

Substituting \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1, results in the characteristic equation:

• \( r^2 - 4r + 4 = 0 \)

This quadratic equation can be solved using factorization or the quadratic formula. Here, it factors neatly as \((r-2)^2 = 0\), indicating a repeated root \( r = 2 \).

The solution for \( r \) directly influences the form of the complementary solution. Different roots would alter the solution's structure, with real, repeated, or complex roots each contributing uniquely to the solution's form.