Question

Solving a First-Order Linear Differential Equation In Exercises \(5-14,\) solve the first-order linear differential equation. $$ y^{\prime}+y \tan x=\sec x $$

Short answer

The solution to the differential equation \(y^{\prime}+y \tan x=\sec x\) is \(y=\cos(x)(\tan(x)+C)\).

Step-by-step solution

Identify the Standard Form

First, rearrange the given differential equation \(y^{\prime}+y \tan x=\sec x\) into the standard form of first-order linear differential equation, which is \(y^{\prime}+P(x)y=Q(x)\). Here, \(P(x)=\tan(x)\) and \(Q(x)=\sec(x)\).

Find the integrating Factor

Next, compute the integrating factor, which is defined as \(e^{\int P(x) dx}\). So, the integrating factor is \(e^{\int \tan(x) dx}\) which simplifies to \(\sec(x)\) because the integral of \(\tan(x)\) is \(-\ln|\cos(x)|\), thus making the integrating factor \(\sec(x)\) or \(1/\cos(x)\).

Multiply the Differential Equation by the Integrating Factor

Now multiply every term in the equation by this integrating factor, yielding \(\sec(x) y^{\prime}+\sec(x) y \tan x=\sec^2x\).

Identify the Left Side as the Derivative of a Product

By carefully observing, the left-hand side of the equation can be written as the derivative of the product of \(\sec(x)\) and \(y\), i.e. \((\sec(x)y)'\).

Integrate Both Sides

Integrating both sides gives \(\sec(x)y=\int \sec^2(x) dx\). The integral of \(\sec^2(x)\) is \(\tan(x)\), so when we integrate, we get \(\sec(x)y=\tan(x)+C\).

Solve for y

To find y as a function of x, we divide both sides by \(\sec(x)\), to get \(y=\cos(x)(\tan(x)+C)\), which is our final solution.

Key concepts

Integrating Factor

Understanding the concept of an integrating factor is crucial when solving first-order linear differential equations. This mathematical tool helps transform a non-exact differential equation into an exact one, which can then be integrated with respect to one variable.

In the context of our example, the integrating factor is derived from the function multiplied by y in the standard form, which is represented as \(e^{\int P(x) dx}\). For the equation given, \(P(x)=\tan(x)\), leading to the integrating factor \(\sec(x)\), as the antiderivative of \(\tan(x)\) is \(\ln|\cos(x)|\), but don't forget the negative sign! Hence, \(e^{\int \tan(x) dx} = e^{-\ln|\cos(x)|} = \sec(x)\).

Applying the integrating factor to each term in the equation simplifies the process substantially, enabling us to combine terms and bring the equation to a form that is easier to handle. This step is pivotal for both achieving the correct solution and understanding the underlying structure of these types of differential equations.

Standard Form of Differential Equation

The standard form of a first-order linear differential equation is essential to identify so that we can apply the technique of finding an integrating factor. It is generally expressed as \(y' + P(x)y = Q(x)\), where \(y'\) is the derivative of y with respect to x, \(P(x)\) is a function of x, and \(Q(x)\) is the source or inhomogeneous term.

In our exercise, the equation \(y'+y\tan(x)=\sec(x)\) is already in standard form with \(P(x) = \tan(x)\) and \(Q(x) = \sec(x)\). Recognizing this allows us to systematically approach the problem using the integrating factor method. Without converting to this form, solving the differential equation could become an insurmountable task.

Derivative of a Product

The derivative of a product of two functions is given by the product rule in calculus. This rule states that if you have two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\).

This concept is applied in Step 4 of our solution to recognize that multiplying y by the integrating factor \(\sec(x)\) and differentiating the result gives us exactly what's on the left side of our modified equation: \(\sec(x) y' + \sec(x)y\tan(x) = (\sec(x)y)'\). The realization that the left side represents the derivative of a product allows us to integrate both sides easily. Without understanding the connection between the differential equation and the derivative of a product, it would be much more challenging to find the solution.

Integral of Trigonometric Functions

The integration of trigonometric functions is a fundamental concept in calculus and plays a key role in solving differential equations. When integrating functions like \(\sec^2(x)\), one must recall the antiderivatives of common trigonometric functions. For example, the integral of \(\sec^2(x)\) that we encounter in Step 5 of our solution is \(\tan(x)\), as per the standard trigonometric integrals.

Hence, when we integrate the right side of our equation, we use the integral \(\int \sec^2(x) dx = \tan(x)\), resulting in \(\sec(x)y = \tan(x) + C\) upon integration where C represents the constant of integration. Mastering the integrals of these functions is instrumental in progressing through calculus and efficiently tackling problems involving trigonometric integrals.