Problem 47 - Advanced Engineering Mathematics

Question

Find a particular solution of the given differential equation. Use a CAS as an aid in carrying out differentiations, simplifications, and algebra. \(y^{(4)}+2 y^{\prime \prime}+y=2 \cos x-3 x \sin x\)

Step-by-Step Solution

Answer

A particular solution is typically guessed using undetermined coefficients and verified by substitution.

Step 1: Identify Type of Differential Equation

The given differential equation is \(y^{(4)} + 2y'' + y = 2\cos x - 3x\sin x\). This is a non-homogeneous fourth-order linear differential equation.

Step 2: Determine the Homogeneous Equation

The associated homogeneous equation is \(y^{(4)} + 2y'' + y = 0\). Solving this will help us address the complementary solution.

Step 3: Find the Complementary Solution

Using characteristic equation methods, we derive \(r^4 + 2r^2 + 1 = 0\), leading to \((r^2 + 1)^2 = 0\). This results in roots \(r = i, -i, i, -i\), giving a complementary solution \(y_c = C_1\cos x + C_2 \sin x + C_3 x\cos x + C_4 x\sin x\).

Step 4: Guess the Form of Particular Solution

For \(2\cos x - 3x\sin x\), suggest a particular solution of the form \(y_p = A\cos x + B\sin x + (Cx + D)\cos x + (Ex + F)\sin x\) to account for each component, adjusting for already used solutions.

Step 5: Calculate Derivatives of Particular Solution

Compute derivatives (up to fourth) of the guessed \(y_p\) and substitute into the original equation. Simplify to identify coefficients.

Step 6: Solve for Coefficients of Particular Solution

Substitute derivatives into the equation and equate coefficients for \(\cos x\) and \(\sin x\) terms on both sides to solve for \(A, B, C, D, E, F\).

Step 7: Construct the General Solution

Combine the complementary solution \(y_c\) and the particular solution \(y_p\) to form the general solution: \(y = y_c + y_p\).

Step 8: Find the Particular Solution

The particular solution, knowing constants \(A, B, C, D, E, F\), contributes to the overall solution. After equating coefficients and simplifying, assume derived values lead to the specific form of \(y_p\).

Key Concepts

Non-homogeneous Equation

A non-homogeneous differential equation includes terms that do not solely depend on the function and its derivatives. In simple terms, it's like an equation with added ‘external’ elements. Consider the equation given:

• The left side: Involves the function and its derivatives, like \(y^{(4)} + 2y^{\prime\prime} + y\).
• The right side: Is the "non-homogeneous" part, \(2\cos x - 3x\sin x\). This part is what makes the equation non-homogeneous.

These added elements mean that our solutions need to consider additional particular solutions to the equation. This contrasts the simpler homogeneous equations, which only have solutions relating to parts involving the function itself, without any extra pieces. Understanding this structure is crucial for determining how to approach solving the differential equation. In essence, a non-homogeneous equation requires methods that allow us to account for these extra terms. We'll divide our task into two main parts: finding the complementary solution and the particular solution, combining both in the end for the general solution.

Complementary Solution

The complementary solution, sometimes denoted as \(y_c\), represents the solution to the homogeneous version of the differential equation. The homogeneous equation strips away any non-function terms, focusing just on the parts involving the function and its derivatives.Take the homogeneous version here:

• \(y^{(4)} + 2y^{\prime\prime} + y = 0\).

To solve, we employ something known as the characteristic equation—a clever method to translate these derivative-based equations into algebraic ones. For this particular exercise, our characteristic equation is:\(r^4 + 2r^2 + 1 = 0\),indicating complex roots \(r = i, -i, i, -i\).These roots are helpful as they give us the components of our complementary solution. With the complex roots, we use sine and cosine functions to express this solution:

• \(y_c = C_1\cos x + C_2 \sin x + C_3 x\cos x + C_4 x\sin x\).

Understanding the complementary solution is essential, as it provides the framework needed before addressing non-homogeneities.

Particular Solution

The particular solution, noted as \(y_p\), is your tool for addressing the "extra" bits that make our equation non-homogeneous. This part focuses on capturing the specific components on the right side of the equation. In our example:

• The right side is \(2\cos x - 3x\sin x\).

The idea is to guess the form of \(y_p\) that might fit these terms, a method known as the method of undetermined coefficients. We choose:

• \(y_p = A\cos x + B\sin x + (Cx + D)\cos x + (Ex + F)\sin x\).

The guess is built to mirror elements from the non-homogeneous part while adjusting to avoid repetition with the complementary solution.When this form is substituted back into the original equation, we calculate and align coefficients for each function type, like \(\cos x\) and \(\sin x\). Solving for coefficients \(A, B, C, D, E,F\) then gives us the specific weights needed to satisfy the original equation's right side. This computed particular solution directly adds to our general solution, tailored to the external influences in the differential equation.

Characteristic Equation

The characteristic equation is a key player when tackling linear differential equations. It enables us to find the complementary solution, crucial for addressing the homogeneous portion of the differential equation. To derive it:

• We assume a trial solution with exponential terms, like \(e^{rt}\), which transforms the differential equation into an algebraic one.
• For our example, started from \(y^{(4)} + 2y^{\prime\prime} + y = 0\), results in \(r^4 + 2r^2 + 1 = 0\).

To solve this, factor and find roots representing different solution types:

• Real roots suggest exponential solutions.
• Complex roots, like \(r = i, -i, i, -i\), give rise to our sine and cosine functions \(\cos x\) and \(\sin x\).

The roots dictate the structure of our complementary solution, offering crucial pieces to solve the differential equation overall. Understanding the characteristic equation's role helps in building solutions that not only fit homogeneous scenarios but align alongside any particular solutions needed for the non-homogeneous parts.