Problem 17 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$(\sin x) y^{\prime}-y \cos x=\sin 2 x, \quad y(\pi / 2)=2$$

Step-by-Step Solution

Answer

The short answer to the given initial-value problem is: \(y(x) = 2\sin^2 x\).

Step 1: Identify the type of differential equation and check its linearity

We are given the differential equation: \[(\sin x) y^{\prime}-y \cos x = \sin 2x\] Let's rewrite it in a more standard form: \[y^{\prime} - \frac{\cos x}{\sin x} y = \frac{\sin 2x}{\sin x}\] We can see that it is a first-order linear differential equation, as it has the form: \[y^{\prime} + p(x) y = Q(x)\] where \(p(x) = -\frac{\cos x}{\sin x}\) and \(Q(x) = \frac{\sin 2x}{\sin x}\).

Step 2: Solve the differential equation, finding the general solution

To solve the first-order linear differential equation, we need to find an integrating factor. An integrating factor can be found by taking the exponential of the integral of the function p(x): \[I(x) = e^{\int p(x) dx} = e^{-\int\frac{\cos x}{\sin x} dx}\] To integrate \(\frac{\cos x}{\sin x}\), we can perform substitution: let \(u = \sin x\), then \(du = \cos x dx\). Therefore, the integral becomes: \[-\int \frac{1}{u} du = -\ln|u|\] Now we substitute back for x: \[-\ln|\sin x|\] So, the integrating factor \(I(x) = e^{-\ln|\sin x|} = \frac{1}{|\sin x|}\). We can drop the absolute value since the sign won't matter when multiplying both sides of the equation. Now we multiply both sides of the differential equation by the integrating factor: \[\frac{1}{\sin x}[y^{\prime} - \frac{\cos x}{\sin x} y] = \frac{1}{\sin x}\cdot\frac{\sin 2x}{\sin x}\] Simplifying the equation, we get: \[\frac{1}{\sin x}y^{\prime} - \frac{\cos x}{\sin^2 x} y = 2\cos x\] Notice that the left-hand side is the derivative of \(\frac{y}{\sin x}\): \[\frac{d}{dx}[\frac{y}{\sin x}] = 2\cos x\] Integrating both sides with respect to x: \[\int \frac{d}{dx}[\frac{y}{\sin x}] dx = \int 2\cos x dx\] This simplifies to: \[\frac{y}{\sin x} = 2\sin x + C\] We have found the general solution for the given differential equation.

Step 3: Apply the initial condition to find the particular solution

Now we apply the initial condition \(y(\pi / 2) = 2\) to find the particular solution. Plugging in the values: \[\frac{2}{\sin (\pi / 2)} = 2\sin (\pi / 2) + C\] This simplifies to: \[2 = 2 + C\] \[C = 0\] Therefore, the particular solution is: \[y(x) = 2\sin x\sin x = 2\sin^2 x\] So, we have found the particular solution for the given initial-value problem: \[y(x) = 2\sin^2 x\]

Key Concepts

Integrating Factor

An integrating factor is a tool used to simplify the process of solving first-order linear differential equations. It transforms a differential equation into a form that is easier to integrate directly. The integrating factor, usually denoted as \(I(x)\), is calculated using the function \(p(x)\) from the differential equation in standard form:

• If the equation is \(y^{\prime} + p(x) y = Q(x)\), identify \(p(x)\).
• Calculate \(I(x) = e^{\int p(x) dx}\).

Once you have the integrating factor, multiply every term in the differential equation by \(I(x)\). This process changes the left-hand side into the derivative of a product, allowing direct integration.
In the given exercise, the integrating factor is calculated from \(p(x) = -\frac{\cos x}{\sin x}\). By integrating \(-\ln|\sin x|\) and simplifying, \(I(x)\) is found to be \(\frac{1}{\sin x}\). Applying this integrating factor transforms the equation into one that can be easily integrated to find the solution.

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with a specific condition, allowing for the determination of a unique solution. This condition, provided as an initial value, often specifies the value of the function at a particular point. This is crucial because, without it, differential equations might have infinitely many solutions known as general solutions.
In the context of the given problem, the initial condition is \(y(\pi/2) = 2\). This constraint allows us to determine the particular solution by substituting these values into the general solution. The initial value ensures that the solution not only satisfies the equation but also passes through this specified point in the solution curve.
Knowing about initial-value problems is essential for solving practical problems where specific conditions are expected to hold at the start or at the boundary of the problem. It transforms an otherwise broad solution into a tailor-fit answer.

General Solution

The general solution of a differential equation represents the set of all possible solutions. It includes arbitrary constants, indicating that it covers every possible curve that could satisfy the equation. These arbitrary constants are crucial because they allow the general solution to adapt to different initial or boundary conditions.
In the original exercise, after determining the integrating factor and integrating both sides, the solution is expressed as \(\frac{y}{\sin x} = 2\sin x + C\). Here, \(C\) is the arbitrary constant that represents the family of curves or solutions.
When a specific initial condition, such as \(y(\pi/2) = 2\), is applied, we substitute it into the general solution to solve for \(C\). In this context, \(C = 0\), giving us the particular solution \(y(x) = 2\sin^2 x\). This process isolates the particular solution for this specific situation from the general solution, ensuring it aligns perfectly with the given condition.