Problem 61 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=e^{x}-x+16$$

Step-by-Step Solution

Answer

The solution is \( y(x) = c_1 e^x + c_2 xe^x + c_3 x^2 e^x + \frac{1}{6} x^3 e^x - \frac{1}{6} x^2 - 16 \).

Step 1: Find the Homogeneous Solution

The given differential equation is \( y''' - 3y'' + 3y' - y = 0 \). Start by solving the homogeneous equation. The characteristic equation is \( r^3 - 3r^2 + 3r - 1 = 0 \). This can be factored into \( (r - 1)^3 = 0 \), giving the repeated root \( r = 1 \) with multiplicity 3. Thus, the general solution of the homogeneous equation is \( y_h(x) = c_1 e^x + c_2 xe^x + c_3 x^2 e^x \).

Step 2: Determine the Form for Particular Solution

Since the non-homogeneous part is \( e^x-x+16 \), we need to propose a particular solution, \( y_p(x) \). The terms \( e^x \) and \( -x \) need careful consideration due to their presence in the homogeneous solution. For \( e^x \), propose \( Ax^3 e^x \), and for \(-x\) use \( Bx^2 + Cx \). For the constant 16, include an additional constant \( D \). Thus, the trial solution is \( y_p(x) = Ax^3 e^x + Bx^2 + Cx + D \).

Step 3: Compute Derivatives of Particular Solution

Differentiate the particular solution trial \( y_p(x) = Ax^3 e^x + Bx^2 + Cx + D \) three times:- First derivative: \[ y_p' = 3Ax^2 e^x + Ax^3 e^x + 2Bx + C \]- Second derivative: \[ y_p'' = 6Axe^x + 6Ax^2 e^x + 2B \]- Third derivative: \[ y_p''' = 6A e^x + 18Ax e^x + 6Ax^2 e^x \].

Step 4: Substitute into Differential Equation

Substitute the derivatives of \( y_p(x) \) into the original differential equation:\[ y_p''' - 3y_p'' + 3y_p' - y_p = (6A e^x + 18Ax e^x + 6Ax^2 e^x) - 3(6Axe^x + 6Ax^2 e^x + 2B) + 3(3Ax^2 e^x + Ax^3 e^x + 2Bx + C) - (Ax^3 e^x + Bx^2 + Cx + D) \].Simplify this expression and equate terms to the right side \( e^x - x + 16 \).

Step 5: Solve for Coefficients A, B, C, D

Expand and simplify the expression from Step 4 to group equivalent terms. Compare the coefficients of like terms with \( e^x - x + 16 \):- Coefficient of \( x^3 e^x \) gives: \( A = 0 \)- Coefficient of \( x^2 e^x \) gives: \( 0 = 0 \) (already checked)- Coefficient of \( xe^x \) gives: \( 0 = 0 \) - Coefficient of \( e^x \) gives: \( 6A = 1 \rightarrow A = 1/6 \) (Typo in previous analysis, correct by solving equations stepwise)- Coefficient of \( x^2 \) : Solve \(-3B + 3A = 0 \)- Solving gives \( B = -1/6 \)- For \( x \) term, solve: \( 3B - C = 1 \)- Solving gives \( C = 0 \)- Constant term: \( -D = 16 \) leads to \( D = -16 \)Correct trial with correct coefficients: Verify by substituting back and checking.

Step 6: Write the General Solution

The complete solution of the differential equation is the sum of the homogeneous and particular solutions:\[ y(x) = c_1 e^x + c_2 xe^x + c_3 x^2 e^x + \frac{1}{6} x^3 e^x - \frac{1}{6} x^2 - 16 \].This combines both the solutions derived from the original homogeneous R and the non-homogeneous terms addressed in the particular solution.

Key Concepts

Homogeneous Solution

The homogeneous solution of a differential equation is a fundamental concept in solving linear equations. It stems from setting the differential equation equal to zero. This identifies the solutions that form the complementary functions, or solutions, of the associated homogeneous differential equation.

For the given problem, the homogeneous differential equation is:

• \( y''' - 3y'' + 3y' - y = 0 \)

To find the homogeneous solution, we solve the characteristic equation. In this case, it is \( r^3 - 3r^2 + 3r - 1 = 0 \). By factoring, we find that \((r - 1)^3 = 0\), indicating a repeated root \( r = 1 \) with a multiplicity of 3.

This gives the homogeneous solution as:

• \( y_h(x) = c_1 e^x + c_2 xe^x + c_3 x^2 e^x \)

These terms combine linearly to produce the general form of the solution that will be valid for any shift of basis defined by initial conditions.

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies the non-homogeneous part of the equation. It is constructed to account for terms present in the given, non-zero right side of the equation.

For this specific equation, the right side is \( e^x - x + 16 \). The challenge here is to propose a particular solution, \( y_p(x) \), which considers the types of functions present. Since \( e^x \) and \( -x \) are part of the homogeneous solution, adjustments need to be made to ensure no overlap occurs:

• Propose \( A x^3 e^x \) for \( e^x \)
• Use \( Bx^2 + Cx \) for \(-x\)
• Include \( D \) for the constant term \( 16 \)

Therefore, the trial particular solution is:

• \( y_p(x) = Ax^3 e^x + Bx^2 + Cx + D \)

Substituting this into the differential equation, you'll differentiate it and equate like terms to solve for coefficients \( A, B, C, \) and \( D \). This approach finds a specific solution that "fits" the modifications caused by the non-homogeneous part.

Undetermined Coefficients

The method of undetermined coefficients is a straightforward technique used for linear differential equations with constant coefficients. It involves guessing a form for the particular solution with unknown coefficients that we solve for, based on the differential equation.

In the problem given, after hypothesizing the particular solution \( y_p(x) = Ax^3 e^x + Bx^2 + Cx + D \), derive the necessary derivatives and substitute these back into the original differential equation.

• First, derive the expressions through differentiation.
• Substitute these into the equation and equate like terms from both sides.
• Solve for each coefficient \( A, B, C, \) and \( D \).

Once complete, you will find specific values: \( A = \frac{1}{6}\), \( B = -\frac{1}{6} \), \( C= 0\), \( D=-16 \). This method simplifies finding specific solutions by leveraging patterns of function types and their derivatives, ensuring they conform to the structure required by the differential equation's right-hand side.