Question

In Exercises \(16-24\) find the general solution. $$ y^{\prime}+(2 \sin x \cos x) y=e^{-\sin ^{2} x} $$

Short answer

Question: Find the general solution of the given first-order linear ordinary differential equation (ODE): \(y'(x) + 2\sin{x}\cos{x}y(x) = e^{-\sin^2{x}}\). Answer: The general solution to the given ODE for \(y(x)\) is: $$ y(x) = \frac{\int e^{-\frac{1}{2}\cos{2x}} e^{-\sin^2{x}} dx}{e^{-\frac{1}{2}\cos{2x}}} $$ Explanation: We found the integrating factor, multiplied both sides of the equation by it, and then integrated both sides. Since we could not explicitly integrate the functions, we left the general solution in the form of an implicit equation.

Step-by-step solution

Determine the Integrating Factor

We will find the integrating factor \(e^{\int{p(x)dx}}\): $$ e^{\int{(2\sin{x}\cos{x})dx}}. $$ Since \(2\sin{x}\cos{x} = \sin{2x}\), we will integrate \(\sin{2x}\) with respect to \(x\): $$ \int\sin{2x}dx = -\frac{1}{2}\cos{2x} + C, $$ thus, the integrating factor is: $$ e^{-\frac{1}{2}\cos{2x}}, $$ in this case, we can neglect the constant \(C\).

Multiply ODE with Integrating Factor

Now we multiply both sides of the given ODE by the integrating factor \(e^{-\frac{1}{2}\cos{2x}}\) and simplify: $$ \begin{aligned} &e^{-\frac{1}{2}\cos{2x}}(y'(x)+ (2\sin{x}\cos{x})y(x))\\ & = e^{-\frac{1}{2}\cos{2x}}e^{-\sin^2{x}} \end{aligned} $$

Integrate Both Sides

Now, we integrate both sides of the equation with respect to \(x\): $$ \int e^{-\frac{1}{2}\cos{2x}} (y'(x) + (2\sin{x}\cos{x})y(x)) dx = \int e^{-\frac{1}{2}\cos{2x}} e^{-\sin^2{x}} dx $$ At this point, we cannot explicitly integrate both functions, so we leave them as an implicit integral.

Solve for \(y(x)\)

Solve for \(y(x)\) by dividing both sides by the integrating factor. This gives us the general solution in the form of an implicit equation: $$ y(x) = \frac{\int e^{-\frac{1}{2}\cos{2x}} e^{-\sin^2{x}} dx}{e^{-\frac{1}{2}\cos{2x}}} $$ This is the general solution to the given ODE for \(y(x)\).

Key concepts

Integrating Factor

The concept of an integrating factor is a powerful tool used to solve first-order linear differential equations. It is a function, typically denoted as \(\mu(x)\), which when multiplied by the original differential equation, simplifies it into an easily integrable form. In essence, the integrating factor transforms a non-exact differential equation into an exact one.

For the differential equation presented in the exercise, \(\mu(x)\) is found by exponentiating the integral of the coefficient of \(y\) from the original equation. Given \( p(x) = 2\sin x \cos x \) in our exercise, which simplifies to \(\sin{2x}\), the integrating factor becomes \(e^{-\frac{1}{2}\cos{2x}}\), after integrating \(\sin{2x}\) and applying the exponential function. The absence of the constant \(C\) is justified as it would cancel out during the process of solving the equation.

First-order Differential Equations

First-order differential equations are equations involving the first derivative of the unknown function and the function itself. These equations are fundamental in mathematical modeling of physical processes and are the starting point for understanding higher-order differential equations.

In the given exercise, the differential equation is first-order and linear, having the form \(y^{\' }+p(x)y=g(x)\), with \(p(x)\) being \(2\sin x \cos x\) and \(g(x)\) being \(e^{-\sin ^{2} x}\). The goal with such an equation is to isolate \(y\) and find its general solution, which represents a family of solutions depending on an arbitrary constant due to the process of integration.

Implicit Solutions

An implicit solution to a differential equation involves the function \(y\) and its derivatives, all intermingled in an equation without having an explicitly solved form for \(y\). In other words, unlike explicit solutions where \(y\) is isolated on one side of the equation, implicit solutions present \(y\) within an equation that may not be directly solvable for \(y\).

After applying the integrating factor and integrating both sides of the equation in the given problem, we find that the solution for \(y(x)\) cannot be expressed in a simple form. Hence, the result is left as an implicit integral, representing the general solution of the original differential equation.

Integration Techniques

Integration is a fundamental technique in calculus used to find functions when their rates of change, or derivatives, are known. There are numerous methods of integration, such as substitution, integration by parts, and trigonometric integrals, which can be applied depending on the form of the function we aim to integrate.

In the problem we are considering, direct integration of the terms after applying the integrating factor is not possible due to the complex nature of the functions involved. This illustrates an important point: not all functions are integrable in terms of elementary functions, and sometimes we must express the solution to an integral implicitly or resort to numerical methods for an approximate answer. The technique chosen must align with the integral's complexity and the desired form of the solution.