Question

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

Short answer

The particular solution is \( y = e^{t} \cos t - e^{t} \sin t \).

Step-by-step solution

Differentiate the Proposed Solution

Given the proposed solution function \( y = a e^{t} \, \cos t + b e^{t} \, \sin t \). We need to differentiate \( y \) with respect to \( t \). Use the product rule to find \( y' \), which is: \[ y' = e^{t} \cos t (a - b) + e^{t} \sin t (a + b) \].

Simplify the Derivative

Simplify the derivative expression found in Step 1: it becomes\[ y' = e^{t} ((a - b) \cos t + (a + b) \sin t) \].

Match with Given Equation

Compare the derivative \( y' = e^{t} ((a - b) \cos t + (a + b) \sin t) \) to the given equation \( y^{\prime} = 2 e^{t} \cos t\). Equate the coefficients of \( \cos t \) and \( \sin t \) from both sides. We get two equations: \(a - b = 2\) and \(a + b = 0\).

Solve the System of Equations

Using the equations \( a - b = 2 \) and \( a + b = 0 \), solve for \( a \) and \( b \). Add the equations: \( 2a = 2 \) simplifies to \( a = 1 \). Substitute \( a = 1 \) into \( a + b = 0 \) to find \( b = -1 \).

Write the Particular Solution

Substitute \( a = 1 \) and \( b = -1 \) back into the original form \( y = a e^{t} \cos t + b e^{t} \sin t \). The particular solution to the differential equation is \( y = e^{t} \cos t - e^{t} \sin t \).

Key concepts

Understanding Differential Equations

Differential equations play a vital role in connecting mathematical and real-world problems. They involve functions and their derivatives, describing how quantities change with respect to one another. In our example, we encounter a differential equation where the derivative \( y' \) is defined as \( 2e^{t} \cos t \). This equation illustrates a relationship involving a function's rate of change (here expressed in terms of \( e^{t} \) and \( \cos t \)).

Solving differential equations often means finding a function, such as \( y \), that satisfies the given equation. A particular solution is a specific solution that fits the initial conditions or particular values in the problem. Such solutions are essential to modeling situations accurately.

To find these, we can employ substitutions, compare coefficients, or utilize other calculus methods to identify how functions interrelate and what variables must equal to satisfy the equation.

Using the Product Rule

The product rule is a fundamental tool in calculus used to differentiate functions that are products of two other functions. When you have a function like \( y = a e^{t} \cos t + b e^{t} \sin t \), it consists of terms that are products of a constant, an exponential, and a trigonometric function.

To differentiate each term effectively, apply the product rule, which states that \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( t \). Let's apply it to \( a e^{t} \cos t \):

• Set \( u = a e^{t} \) and \( v = \cos t \).
• Then, \( u' = a e^{t} \) because the derivative of \( e^{t} \) is \( e^{t} \).
• For \( v', \) the derivative of \( \cos t \) is \(-\sin t \).

Combining these, the derivative becomes \( y' = a e^{t}(\cos t - \sin t) + b e^{t}(\sin t + \cos t) \) after applying the product rule again for \( b e^{t} \sin t \). This allows us to follow through with the method to match and solve the differential equation.

Solving a System of Equations

A system of equations arises here as we solve for constants \( a \) and \( b \) matching the derivative with the given differential equation. Solving such a system involves algebraic techniques to determine these constants but can seem daunting without practice.

From \( y' = 2e^{t} \cos t \) and the derived solution, we relate and match terms to form:

• \( a - b = 2 \)
• \( a + b = 0 \)

These linear equations form a system because they are coupled; solving one impacts the other. Here's how you address it:

• Combine them by adding: \( (a - b) + (a + b) = 2 \)
• This simplifies neatly: \( 2a = 2 \) leading to \( a = 1 \).
• Substitute \( a = 1 \) back into either equation, say \( a + b = 0 \), to find \( b = -1 \).

Eventually, applying these constants, you identify the particular solution: \( y = e^{t} \cos t - e^{t} \sin t \), demonstrating problem-solving strategies in applied mathematics.