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 \(y = e^t(\cos t + \sin t)\).

Step-by-step solution

Differentiate the Given Function

To substitute correctly, we first differentiate the given function: \(y = ae^{t} \cos t + be^{t} \sin t\). Use the product rule and the derivatives \(\frac{d}{dt}(e^t) = e^t\), \(\frac{d}{dt}(\cos t) = -\sin t\), and \(\frac{d}{dt}(\sin t) = \cos t\). The derivative is:\[y^{\prime} = a(e^t \cos t)' + b(e^t \sin t)'\]Calculating each:\(a(e^t \cos t)^{\prime} = ae^t \cos t - ae^t \sin t\)\(b(e^t \sin t)^{\prime} = be^t \cos t + be^t \sin t\)Thus, \(y^{\prime} = ae^t \cos t - ae^t \sin t + be^t \cos t + be^t \sin t\).

Combine Like Terms in Derivative

Combine like terms in the expression for \(y^{\prime}\) obtained in Step 1:\[y^{\prime} = (a+b)e^t \cos t + (b-a)e^t \sin t\]

Equate Derived Expression with Given Equation

Set the derivative expression obtained in Step 2 equal to the given equation \(y^{\prime} = 2e^t \cos t\):\[(a+b)e^t \cos t + (b-a)e^t \sin t = 2e^t \cos t\]

Solve the System of Equations

For the equation in Step 3 to hold for all \(t\), coefficients of \(e^t \cos t\) and \(e^t \sin t\) must match. Therefore, we have:1. \(a + b = 2\)2. \(b - a = 0\)Solve this system:From equation (2), we have \(b = a\). Substitute into equation (1):\[a + a = 2\]\[2a = 2\]\[a = 1\]Substitute \(a = 1\) into \(b = a\):\[b = 1\]

Write the Particular Solution

Now that we have values of \(a\) and \(b\), substitute back into the original \(y\) function:\[y = 1e^{t} \cos t + 1e^{t} \sin t = e^t(\cos t + \sin t)\]

Key concepts

Particular Solution

When solving differential equations, finding a particular solution is essential. A particular solution is a specific function that satisfies not only the differential equation itself but also any initial condition or requirements that are given. In this problem, the aim is to find a particular solution for the given differential equation \(y^{\prime} = 2 e^{t} \cos t\).
To achieve this, we substitute a general solution, \(y = ae^{t} \cos t + be^{t} \sin t\), where \(a\) and \(b\) are constants. The process involves calculating the derivative of this function, aligning it with the differential equation, and finding values of \(a\) and \(b\) that make the derivative match the given equation.
Once these values are determined, they are substituted back into the original function \(y\) to obtain a particular solution that meets the specific condition of the problem.

Product Rule

The product rule is a fundamental concept in calculus used to differentiate expressions where two functions are multiplied together. When you have a function such as \(u(t) \cdot v(t)\), its derivative is given by \((u)^{\prime}v + u(v)^{\prime}\). The given problem requires using the product rule because each term of the function \(y = ae^{t} \cos t + be^{t} \sin t\) is a product of two simpler functions.

The derivative of \(y\) is determined by differentiating each term separately using the product rule:

• For \(ae^{t} \cos t\), it becomes \(a(e^t \cos t)' = ae^t \cos t - ae^t \sin t\).
• For \(be^{t} \sin t\), it becomes \(b(e^t \sin t)' = be^t \cos t + be^t \sin t\).

This systematic approach allows for evaluating the original differential equation accurately by breaking it into manageable parts. Understanding and applying the product rule effectively is crucial in solving many differential equations.

System of Equations

The solution to the original differential equation requires solving a system of equations. When you compare the differentiated function to the given equation \(y^{\prime} = 2e^t \cos t\), you need the coefficients of \(e^t \cos t\) and \(e^t \sin t\) in the derived expression to match the original differential equation.
The comparison leads to two equations:

• Equation 1: \(a + b = 2\)
• Equation 2: \(b - a = 0\)

These equations form a system that can be solved to find \(a\) and \(b\). Solving such systems involves:

• Substituting \(b = a\) derived from Equation 2 into Equation 1
• Calculating \(a = 1\), and consequently \(b = 1\).

By solving this system, the specific values of \(a\) and \(b\) are found, which are then substituted back into the original function to complete the solution. Solving systems of equations is a vital technique in various mathematical problems, ensuring that all conditions are consistently met.