Problem 49 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}+6 y^{\prime}+9 y=-x e^{4 x}$$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 e^{-3x} + C_2 x e^{-3x} + \left( -\frac{1}{19}x + \frac{4}{57} \right)e^{4x} \).

Step 1: Identify the Homogeneous Equation

First, consider the associated homogeneous equation, which is obtained by setting the right-hand side to zero: \[ y'' + 6y' + 9y = 0. \]

Step 2: Solve the Homogeneous Equation

The characteristic equation for the homogeneous equation is:\[ r^2 + 6r + 9 = 0. \]Solving this, we find the roots are \( r = -3 \) with multiplicity 2, so the general solution of the homogeneous equation is:\[ y_h = C_1 e^{-3x} + C_2 x e^{-3x}. \]

Step 3: Formulate Particular Solution Using Method of Undetermined Coefficients

Now, we guess a particular solution \( y_p \) based on the form of the non-homogeneous term \(-x e^{4x}\). We try:\[ y_p = (Ax + B)e^{4x}. \]However, note that since \( e^{4x} \) is not a solution of the homogeneous equation, we don't need extra multiplicative factors of \( x \).

Step 4: Differentiate Particular Solution

Next, find the first and second derivatives of the particular solution:\[ y_p' = (A + 4Ax + 4B)e^{4x}, \]\[ y_p'' = (8A + 4Ax) e^{4x} + 16B e^{4x}. \]

Step 5: Substitute into Original Differential Equation

Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the left side of the original differential equation:\[ (8A + 4Ax) e^{4x} + 16B e^{4x} + 6(A + 4Ax + 4B) e^{4x} + 9(Ax + B)e^{4x} = -x e^{4x}. \]

Step 6: Combine Like Terms and Solve for Coefficients

After expansion, combine like terms:\[ (4A + 6A + 9A)x e^{4x} + (8A + 24B + 6B) e^{4x} = -x e^{4x}. \] This simplifies to:\[ (19A)x e^{4x} + (32A + 24B)e^{4x} = -xe^{4x}. \]Set coefficients equal to solve:\[ 19A = -1, \]\[ 32A + 24B = 0. \]

Step 7: Solve System of Equations for A and B

Solving \( 19A = -1 \), we find \( A = -\frac{1}{19} \).Using \( 32A + 24B = 0 \), substitute \( A \):\[ 32\left(-\frac{1}{19}\right) + 24B = 0 \]\[ -\frac{32}{19} + 24B = 0 \]\[ 24B = \frac{32}{19} \]\[ B = \frac{32}{19 \times 24} = \frac{4}{57}. \]

Step 8: Write Particular Solution

The particular solution is now:\[ y_p = \left( -\frac{1}{19}x + \frac{4}{57} \right)e^{4x}. \]

Step 9: Write General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:\[ y = y_h + y_p = C_1 e^{-3x} + C_2 x e^{-3x} + \left( -\frac{1}{19}x + \frac{4}{57} \right)e^{4x}. \]

Key Concepts

Homogeneous Equation

When working with differential equations, a fundamental step is to first consider the homogeneous equation. This involves setting the non-homogeneous part (the right side of the equation) to zero. For the given equation:
\[ y'' + 6y' + 9y = -x e^{4x} \]
We remove the non-essential component, leading to:
\[ y'' + 6y' + 9y = 0 \]
This is the homogeneous equation. It helps us find the complementary solution, which is critical in building the solution to the entire problem. The essence of the homogeneous part is to evaluate the behavior of solutions based solely on the differential operator.

Characteristic Equation

Once you've identified the homogeneous equation, the next step is to derive the characteristic equation. This is done by replacing the derivatives in the homogeneous equation with powers of \( r \). The equation\[ y'' + 6y' + 9y = 0 \]translates to
\[ r^2 + 6r + 9 = 0 \]
in terms of \( r \). This quadratic equation solves easily by factoring or using the quadratic formula.

• Factoring typically gives the roots, which indicate the type of solution: real, repeated, or complex.
• Here, the roots are both \( r = -3 \), signifying a repeated root.

The general solution formed by these roots captures all solutions possible for the homogeneous equation and sets us up to solve for what remains: the particular solution.

Undetermined Coefficients

The method of undetermined coefficients involves making an educated guess about the form of the particular solution, \( y_p \). This technique is particularly valuable when dealing with linear differential equations with constant coefficients and is well-suited for when the non-homogeneous term follows specific patterns.

• Look at the non-homogeneous part, \(-x e^{4x}\), and formulate a trial solution that resembles it.
• For our problem, the guess is: \( y_p = (Ax + B)e^{4x} \).

Next, we differentiate the trial solution, plug it back into the differential equation, and solve for the unknown coefficients \( A \) and \( B \). This step ensures the particular solution effectively accounts for the non-homogeneous portion of the equation.

Particular Solution

Having guessed a form for the particular solution, our new task is finding the coefficients that make this guess satisfy the original differential equation.

• Differentiation of the guessed solution provides expressions for \( y_p' \) and \( y_p'' \).
• These are substituted back into the complete differential equation:\[ y^{ ext{original}}\]
• By equating coefficients of like terms on both sides, we establish equations for \( A \) and \( B \).

Root these values to finally determine:
\[ y_p = \left( -\frac{1}{19}x + \frac{4}{57} \right)e^{4x} \]
With the particular solution in hand, the general solution combines it with the homogeneous solution, painting a full picture of all possible solutions: \( y = y_h + y_p \), providing a comprehensive answer.