Question

Substitute \(y=a \cos (2 t)+b \sin (2 t)\) into \(y^{\prime}+y=4 \sin (2 t)\) to find a particular solution.

Short answer

The particular solution is \( y = -\frac{8}{5}\cos(2t) + \frac{4}{5}\sin(2t) \).

Step-by-step solution

Differentiate the given function

First, determine the derivative of the function \( y = a \cos(2t) + b \sin(2t) \). Use the chain rule to find \( y' \). The derivative of \( \cos(2t) \) is \( -2\sin(2t) \) and the derivative of \( \sin(2t) \) is \( 2\cos(2t) \). So, \( y' = -2a\sin(2t) + 2b\cos(2t) \).

Substitute into differential equation

Substitute \( y \) and \( y' \) into the differential equation \( y' + y = 4\sin(2t) \). This gives: \( -2a\sin(2t) + 2b\cos(2t) + a\cos(2t) + b\sin(2t) = 4\sin(2t) \).

Rearrange and combine terms

Combine like terms in the equation \( -2a\sin(2t) + b\sin(2t) + 2b\cos(2t) + a\cos(2t) = 4\sin(2t) \), resulting in \((a + 2b)\cos(2t) + (b - 2a)\sin(2t) = 4\sin(2t) \).

Solve the system of equations

To satisfy the equation for all \( t \), equate the coefficients of \( \sin(2t) \) and \( \cos(2t) \) to those in \( 0\cos(2t) + 4\sin(2t) \). Thus, \( a + 2b = 0 \) and \( b - 2a = 4 \). Solve these equations simultaneously to find \( a \) and \( b \). Solving the first equation, \( a = -2b \). Substitute \( a = -2b \) into the second equation: \( b - 2(-2b) = 4 \). This becomes \( 5b = 4 \), giving \( b = \frac{4}{5} \). Substitute \( b = \frac{4}{5} \) back into \( a = -2b \), yielding \( a = -\frac{8}{5} \).

Write the particular solution

With \( a = -\frac{8}{5} \) and \( b = \frac{4}{5} \), substitute back into \( y = a\cos(2t) + b\sin(2t) \) to get the particular solution: \( y = -\frac{8}{5}\cos(2t) + \frac{4}{5}\sin(2t) \).

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. It serves to establish relationships between changing quantities. Differential equations are fundamental in numerous scientific disciplines because they can describe real-world phenomena such as population dynamics, heat conduction, wave propagation, and more.
In the exercise, we are dealing with a first-order linear differential equation of the form \( y' + y = 4 \sin(2t) \). Here, \( y' \) represents the derivative of \( y \) with respect to \( t \), and \( y \) represents the function itself. The right side of the equation, \( 4 \sin(2t) \), is a forcing function that drives the dynamic behavior of the system described by the differential equation.
To solve such an equation, the common approach is to find a particular solution that satisfies the equation. A particular solution is not the only solution but one specific solution that satisfies the given equation under specific conditions. The process involves finding values for any constants in your potential solutions.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play pivotal roles in a variety of mathematical contexts, especially in this exercise. These functions, \( \sin(2t) \) and \( \cos(2t) \), relate to the angles and ratios in right-angled triangles, which can also represent periodic phenomena such as sound waves and alternating current.
In the context of our differential equation, \( y = a \cos(2t) + b \sin(2t) \), where \( a \) and \( b \) are constants, provides a way to express the solution using these trigonometric functions. Each component describes its own wave-like behavior, thus making it suitable to solve the equation.
Differentiating these trigonometric expressions was a crucial step in forming the derivative \( y' \) needed for the substitution into the differential equation. The properties of trigonometric derivatives, such as \( \frac{d}{dt}[\cos(2t)] = -2\sin(2t) \) and \( \frac{d}{dt}[\sin(2t)] = 2\cos(2t) \), were instrumental to this process.

Substitution Method

The substitution method is a powerful tool in solving differential equations. It involves substituting guessed or proposed solutions into the original differential equation. It allows for simplifying the process of verifying if the proposed solution works. In our case, the substitution involved plugging the expressions for \( y \) and its derivative \( y' \) back into the original equation \( y' + y = 4 \sin(2t) \).
Substituting these into the equation allows us to rearrange and manage complex terms involving \( \sin(2t) \) and \( \cos(2t) \). We then equate the coefficients of same trigonometric terms (like \( \sin(2t) \) and \( \cos(2t) \)) on both sides to find correct values for \( a \) and \( b \), ensuring the solution satisfies the differential equation for all \( t \).
Substitution simplifies finding the values of unknown coefficients, streamlining the solution path through systematic algebraic manipulation and valuable insights into the symmetry and structure of the functions involved.