Problem 44 - Advanced Engineering Mathematics

Question

Discuss why it makes intuitive sense to presume that the linear differential equation \(y^{\prime \prime}+2 y^{\prime}+4 y=5 \sin t\) has a solution of the form \(y=A \sin t+B \cos t\), where \(A\) and \(B\) are constants. Then find specific constants \(A\) and \(B\) so that \(y=A \sin t+B \cos t\) is a particular solution of the \(\mathrm{DE}\).

Step-by-Step Solution

Answer

The solution form is intuitive due to the trigonometric non-homogeneous term; the specific solution is \( y = \frac{5}{3} \sin t - \frac{5}{9} \cos t \).

Step 1: Analyze the Nature of the Equation

The given differential equation is \( y'' + 2y' + 4y = 5\sin t \). It is a second-order linear differential equation with constant coefficients. The non-homogeneous term is \( 5\sin t \). A plausible solution form for the steady state or particular solution of such an equation is often related to the form of the non-homogeneous term.

Step 2: Propose a Solution Form

Since the non-homogeneous term is \( 5\sin t \), we assume a particular solution of the form \( y = A \sin t + B \cos t \). This form is chosen because sines and cosines are solutions of linear differential equations involving trigonometric functions as non-homogeneous terms.

Step 3: Differentiate the Proposed Solution

Find the first and second derivatives of the proposed solution: \( y = A \sin t + B \cos t \). \[ y' = A \cos t - B \sin t \] \[ y'' = -A \sin t - B \cos t \]

Step 4: Substitute into the Differential Equation

Substitute \( y, y' \), and \( y'' \) back into the original differential equation:\[(-A \sin t - B \cos t) + 2(A \cos t - B \sin t) + 4(A \sin t + B \cos t) = 5 \sin t\]

Step 5: Simplify the Equation

Combine like terms, separating \( \sin t \) and \( \cos t \) terms:\[(-A + 4A) \sin t + (-B + 2A + 4B) \cos t + 2(-B) \sin t = 5 \sin t\] This simplifies to: \[3A \sin t + (6B + 2A) \cos t = 5 \sin t\]

Step 6: Solve for Coefficients

Equate coefficients of \( \sin t \) and \( \cos t \) separately:1. \( 3A = 5 \) which implies \( A = \frac{5}{3} \)2. \( 6B + 2A = 0 \) which implies \( B = -\frac{5}{9} \).

Step 7: Verify the Solution

Substitute back \( A = \frac{5}{3} \) and \( B = -\frac{5}{9} \) into \( y = A \sin t + B \cos t \) and check the original differential equation to verify the solution.

Key Concepts

Non-Homogeneous Differential Equation

A non-homogeneous differential equation is a type of differential equation that includes a non-zero term, usually called the non-homogeneous term, on its right-hand side. Unlike homogeneous equations, which equate to zero, these equations are of the form \( y'' + p(t) y' + q(t) y = g(t) \). Here, \( g(t) \) is not zero and represents the non-homogeneous part.

In our specific problem, the equation is \( y'' + 2y' + 4y = 5\sin t \). The term \( 5\sin t \) is the non-homogeneous element. This changes how we approach finding the solution.

• We must incorporate this term when guessing our particular solution form.
• Non-homogeneous differential equations typically need a complementary (or homogeneous) solution and a particular solution.

Understanding the type of differential equation helps guide the selection of a solution approach and form.

Particular Solution

A particular solution is a solution to a non-homogeneous differential equation that specifically addresses the non-homogeneous part. It doesn't encompass the full family of solutions but provides a specific function that satisfies the entire equation, including the non-zero term.

The process for finding a particular solution begins with hypothesizing a form based on the non-homogeneous term. In this case, since the term is \( 5\sin t \), we choose the solution form \( y = A \sin t + B \cos t \). This form leverages the properties of trigonometric functions in differential equations.

Once the form is proposed:

• Calculate the first and second derivatives.
• Substitute back into the original equation.
• Equate like terms to solve for constants \( A \) and \( B \).

This method ensures that our particular solution fits the unique characteristics of the given non-homogeneous equation.

Constant Coefficients

Constant coefficients refer to the coefficients in a differential equation that are constant values—meaning they do not change with respect to the variable. They simplify the analysis and solution of differential equations by allowing certain assumptions about the solution forms.

In our given equation, \( y'' + 2y' + 4y = 5\sin t \), the coefficients 2 and 4 are constants. Constant coefficients allow us to consider standard forms like \( e^{rt} \), \( sin(t) \), and \( cos(t) \) when trying to find solutions.

• These solutions utilize known characteristic equations for constant coefficients to derive general forms.
• The trigonometric approach for the particular solution also ties back to these constant elements.

By working with constant coefficients, it becomes easier to apply established solution methods.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are particularly useful when dealing with differential equations with non-homogeneous terms that are also trigonometric in nature. They help in forming solutions that appropriately match and counterbalance the non-homogeneous parts.

In the context of our problem, the function \( 5\sin t \) suggests using a specific trigonometric solution form: \( y = A \sin t + B \cos t \). This choice is intuitive because:

• The derivatives of sine and cosine functions cycle among themselves.
• These functions accommodate solutions for differential equations with sinusoidal non-homogeneous terms.

Trigonometric functions are useful tools for modeling periodic behavior and are therefore ideal for tackling non-homogeneous differential equations with similar characteristics.