Question

Find the solution to the indicated initial value problem, and use ezplot to plot it. \(y^{\prime}=-y+\cos t\) with \(y(0)=2\) over \([0,5]\).

Short answer

The solution is \( y(t) = \frac{\cos t + \sin t}{2} + \frac{3}{2} e^{-t} \) with \( y(0)=2 \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' = -y + \cos t \) with the initial condition \( y(0) = 2 \). This is a first-order linear ordinary differential equation.

Find the Integrating Factor

To solve this equation, we first find the integrating factor. The standard form is \( y' + P(t)y = Q(t) \) where \( P(t) = 1 \) and \( Q(t) = \cos t \). The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^t \).

Multiply through by the Integrating Factor

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

Write the Left Side as a Derivative

Notice that the left side of the equation is now the derivative of \( e^t y \). Therefore, we can rewrite it as:\[ \frac{d}{dt}(e^t y) = e^t \cos t. \]

Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^t y) \, dt = \int e^t \cos t \, dt. \]The left side becomes \( e^t y \). The right side requires the use of integration by parts.

Solve the Integral by Parts

Using integration by parts, where \( u = \cos t \) and \( dv = e^t \, dt \), we find:\[ \int e^t \cos t \, dt = e^t \cos t + e^t \sin t - \int e^t \cos t \, dt. \]Solve this to find the particular integral by rearranging terms. After simplification, the integral becomes:\[ \int e^t \cos t \, dt = \frac{e^t (\cos t + \sin t)}{2}. \]

Solve for y(t)

Substitute back to find \( y(t) \):\[ e^t y = \frac{e^t (\cos t + \sin t)}{2} + C. \]Solve for \( y(t) \):\[ y(t) = \frac{\cos t + \sin t}{2} + Ce^{-t}. \]

Use Initial Condition to Solve for C

Use the initial condition \( y(0) = 2 \) to find \( C \):\[ 2 = \frac{1}{2} + C \quad \Rightarrow \quad C = \frac{3}{2}. \]

Write the Final Solution

Substitute \( C = \frac{3}{2} \) back into \( y(t) \):\[ y(t) = \frac{\cos t + \sin t}{2} + \frac{3}{2} e^{-t}. \]

Plot the Solution

Use a software tool like MATLAB's `ezplot` function to plot the solution \( y(t) = \frac{\cos t + \sin t}{2} + \frac{3}{2} e^{-t} \) over the interval \([0, 5]\).

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with a specific condition called an initial condition, which helps to find a unique solution. This initial condition is given at the start of the problem. For example, in the exercise above, the initial condition is given by \( y(0) = 2 \).

The process involves solving the differential equation while ensuring that the solution satisfies this initial condition. This is important as it provides the exact solution, instead of a general one that might include arbitrary constants. By applying the initial value, you determine these constants, pinpointing the specific solution needed.

• An IVP must include:
• A differential equation.
• An initial condition, which is usually a value for the function at a certain point.

Integration by Parts

Integration by parts is a helpful technique for finding the integral of a product of functions. It is derived from the product rule of differentiation and is especially useful when dealing with integrals involving products of polynomials, exponentials, or trigonometric functions.

The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

In solving the problem given, integration by parts is used to integrate the right-hand side equation \( \int e^t \cos t \, dt \). By choosing \( u = \cos t \) and \( dv = e^t \, dt \), and applying the formula, we simplify the process of integration and find the particular integral needed to solve the differential equation.

First-Order Linear Differential Equation

A first-order linear differential equation is characterized by having a first derivative and no higher order derivatives. The general form of such an equation is:

• \( y' + P(t)y = Q(t) \)

Where \( P(t) \) and \( Q(t) \) are functions of \( t \). The equation in this exercise, \( y' = -y + \cos t \), can be rearranged into this standard form to identify \( P(t) = 1 \) and \( Q(t) = \cos t \).

These equations are linear because the unknown function and its derivative appear only to the first degree. Solving them typically involves finding an integrating factor, a step that effectively simplifies the equation into one that can be easily integrated.

Integrating Factor

The integrating factor is a technique used to solve first-order linear differential equations. It simplifies these equations by transforming them into a form that can easily be integrated.

For a differential equation of the form \( y' + P(t)y = Q(t) \), the integrating factor \( \mu(t) \) is given by the exponential:

• \( \mu(t) = e^{\int P(t) \, dt} \)

In the example exercise, the integrating factor is \( e^t \) since \( P(t) = 1 \). Multiplying the entire equation by this factor transforms it, allowing you to express the left side as the derivative of a product. This crucial simplification makes it possible to integrate both sides directly, leading to the solution.

The integrating factor effectively "balances out" the equation, handling complexities that might otherwise make integration directly challenging. It is a key tool in solving linear differential equations.