Question

$$y^{\prime}=2 \cos x-y$$

Short answer

The general solution to the differential equation \(y' = 2 \cos(x) - y\) is \(y = 2\cos(x) + 2\sin(x) + Ce^{x}\).

Step-by-step solution

Identify the equation type

Examine the equation to determine its type. The given equation is a first-order linear differential equation of the form \( y' + p(x)y = g(x) \), where \( p(x) = -1 \) and \( g(x) = 2\cos(x) \).

Find the integrating factor

The integrating factor \( \mu(x) \) is found using the formula \( \mu(x) = e^{\int p(x) dx} \). In this case, \( \mu(x) = e^{\int -1 dx} = e^{-x} \).

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor to get the equation in the form \( \mu(x)y' + \mu(x)p(x)y = \mu(x)g(x) \), which becomes \( e^{-x}y' - e^{-x}y = 2e^{-x}\cos(x) \).

Rewrite the left side as the derivative of a product

Notice that the left side of the equation can be rewritten as the derivative of a product of functions \( \mu(x) \) and \( y \), which is \( (e^{-x}y)' \).

Integrate both sides

Integrate both sides of the equation with respect to \( x \) to find \( y \). The integral of the left side is \( e^{-x}y \), and the integral of the right side is \( \int 2e^{-x}\cos(x) dx \), which can be solved by integration by parts or looking up the integral in a table.

Solve the integral of the right side

Using either a table or integration by parts (which may require using it twice), find the integral \( \int 2e^{-x}\cos(x) dx \). This is a typical integral that leads to \( 2e^{-x}\cos(x) + 2e^{-x}\sin(x) + C \) after applying integration by parts. The constant \( C \) will be the constant of integration.

Express the general solution

Put together the results from integration to write down the general solution of the differential equation, which is \( y = e^{x} \left(2e^{-x}\cos(x) + 2e^{-x}\sin(x) + C\right) \), or simplified to \( y = 2\cos(x) + 2\sin(x) + Ce^{x} \).

Key concepts

Integrating Factor

An integrating factor is a powerful tool used to solve first-order linear differential equations, such as the one given in the exercise \( y' = 2 \cos x - y \). The purpose of an integrating factor is to convert a non-exact differential equation into an exact equation, which can then be easily integrated.

The formula to find an integrating factor \( \mu(x) \) is \( \mu(x) = e^{\int p(x) \, dx} \), where \( p(x) \) is a function of \( x \) that multiplies the \( y \) term in the standard form \( y' + p(x)y = g(x) \). In the exercise, \( p(x) = -1 \), so the integrating factor is \( \mu(x) = e^{-x} \). By using the integrating factor, the equation is transformed and ready for the next step, which typically involves integration.

Multiplying through the differential equation by the integrating factor helps us to rewrite the equation in a way that the left-hand side becomes the derivative of a product, allowing for a straightforward integration. The use of integrating factors simplifies the process of finding the general solution to the differential equation, a vital concept for students to master.

Differential Equations

Differential equations, such as \( y' = 2 \cos x - y \), are mathematical equations that relate some function with its derivatives. These types of equations are pivotal in various scientific fields as they describe phenomena where change is constant and can be related to rates at which other changes occur.

First-order linear differential equations, to which the given exercise belongs, are written in the form \( y' + p(x)y = g(x) \). In our example, \( p(x) = -1 \) and \( g(x) = 2\cos(x) \). They are characterized by having only the first derivative of \( y \) and no higher derivatives. The standard approach to solving them is to identify the integrating factor, multiply it through the equation, and perform integration to find the solution.

The beauty of differential equations lies in their ability to model real-world situations with accuracy. Therefore, understanding how to solve these equations is not just a technique but also a way to interpret and predict the behavior of dynamic systems in nature, engineering, economics, and beyond.

Integration by Parts

Integration by parts is a technique used to integrate products of functions and is based on the product rule for differentiation. When encountering an integral that is difficult to evaluate directly, like \( \int 2e^{-x}\cos(x) dx \), integration by parts can be particularly useful.

The formula for integration by parts is derived from the product rule and is given by \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are parts of the original integrand selected by the integrator. Choosing \( u \) and \( dv \) strategically is key to simplifying the integral.

In the context of the exercise, integration by parts is applied to the right-hand side after multiplying through by the integrating factor. In some cases, as noted in the solution, it may even be necessary to apply integration by parts more than once to arrive at a solvable integral. This technique is particularly relevant for a variety of integrals involving exponential functions, trigonometric functions, and products thereof, which frequently appear in differential equations.