Question

\(y^{\prime}+y=\cos t\), with \(y(0)=1\), on \([0,5 \pi]\).

Short answer

The solution is \( y = \frac{1}{2}(\sin t + \cos t) + \frac{1}{2}e^{-t} \) valid on \([0, 5\pi]\).

Step-by-step solution

Recognize the Type of Differential Equation

The given differential equation \( y' + y = \cos t \) is a first-order linear ordinary differential equation. It is in the form \( y' + p(t)y = g(t) \), where \( p(t) = 1 \) and \( g(t) = \cos t \).

Find the Integrating Factor

The integrating factor for a first-order linear ordinary differential equation is given by \( \mu(t) = e^{\int p(t)\, dt} = e^{\int 1 \, dt} = e^t \).

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation \( y' + y = \cos t \) by the integrating factor \( e^t \) to get \( e^t y' + e^t y = e^t \cos t \).

Recognize the Left Side as a Product Derivative

The left-hand side of the equation \( e^t y' + e^t y \) can be written as the derivative of the product \( \frac{d}{dt}(e^t y) \). Thus, the equation becomes \( \frac{d}{dt}(e^t y) = e^t \cos t \).

Integrate Both Sides

Integrate both sides with respect to \( t \):\[ e^t y = \int e^t \cos t \, dt + C \]This involves integration by parts, where \( u = \cos t \) and \( dv = e^t \, dt \). After completing the integration and using tabular integration due to repeating pattern:\[ \int e^t \cos t \, dt = \frac{1}{2}(e^t \sin t + e^t \cos t) \]

Include the Constant of Integration

After integrating, we have:\[ e^t y = \frac{1}{2}(e^t \sin t + e^t \cos t) + C \]

Solve for \( y \)

To solve for \( y \), divide through by \( e^t \):\[ y = \frac{1}{2}(\sin t + \cos t) + C e^{-t} \]

Apply the Initial Condition

Use the initial condition \( y(0) = 1 \) to solve for \( C \).\[ 1 = \frac{1}{2}(\sin 0 + \cos 0) + C e^{0} \]\[ 1 = \frac{1}{2}(0 + 1) + C \]\[ C = \frac{1}{2} \]

Write the Particular Solution

Substitute \( C = \frac{1}{2} \) back into the expression for \( y \):\[ y = \frac{1}{2}(\sin t + \cos t) + \frac{1}{2}e^{-t} \]

Check the Interval of Validity

The problem asks for the solution on \([0, 5\pi]\). Since there are no discontinuities or singularities, the solution \( y = \frac{1}{2}(\sin t + \cos t) + \frac{1}{2}e^{-t} \) is valid over this interval.

Key concepts

Integrating Factor

The integrating factor is an essential tool used to solve first-order linear differential equations efficiently. When you see a differential equation of the form \( y' + p(t)y = g(t) \), the integrating factor \( \mu(t) \) simplifies solving this equation. It is specifically calculated as \( \mu(t) = e^{\int p(t) \, dt} \).
For our differential equation, \( y' + y = \cos t \), we recognize \( p(t) = 1 \). Hence, the integrating factor is \( \mu(t) = e^{\int 1 \, dt} = e^t \).
Multiplying the whole equation by this integrating factor transforms the left-hand side into the derivative of a product, which simplifies further manipulation and integration.

Initial Value Problem

An initial value problem (IVP) combines a differential equation with a specific initial condition, such as \( y(0) = 1 \) in this case. It helps pinpoint the exact solution of a differential equation by using given data about the system’s state at a particular time.
To solve an IVP, follow these steps:

• Identify the differential equation and verify its type.
• Solve the differential equation generally, considering arbitrary constants.
• Apply the initial conditions to determine the unknown constants, thus obtaining a particular solution.

In our scenario, the integration, along with the condition \( y(0) = 1 \), guides us to find the constant \( C \), completing the solution to the problem.

First-Order Linear Differential Equation

A first-order linear differential equation contains only the first derivative of the function and can generally be expressed in the form \( y' + p(t)y = g(t) \). The challenge lies in finding solutions that satisfy the equation along with any initial conditions provided.
Once identified, the use of integrating factors, as described earlier, enables us to transform such equations into simpler forms, making them easier to integrate and thus solve.
In our case, recognizing \( y' + y = \cos t \) as a first-order linear differential equation allowed us to apply these techniques effectively and reduced the complexity of solving the equation.

Solution Validity

Determining the validity of a solution involves checking if the solution satisfies the differential equation across the specified interval. In this exercise, the problem requires a solution over \([0, 5\pi]\).
We found \( y = \frac{1}{2}(\sin t + \cos t) + \frac{1}{2}e^{-t} \) as the solution considering the initial condition. To ensure that it is valid:

• Verify there are no singularities or discontinuities in the solution.
• Confirm the solution fulfills the differential equation and initial condition within the interval.

In this specific problem, the function remains continuous and differentiable across the entire interval \([0, 5\pi]\), ensuring the solution's validity.