Chapter 1: Problem 44
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 \(\mathrm{DE}\).
Short Answer
Step by step solution
Analyze the Nature of the Equation
Propose a Solution Form
Differentiate the Proposed Solution
Substitute into the Differential Equation
Simplify the Equation
Solve for Coefficients
Verify the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-Homogeneous Differential Equation
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.
Particular Solution
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 \( B \).
Constant Coefficients
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.
Trigonometric Functions
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.