Question

Solution of \((\mathrm{y} / \mathrm{x}) \cos (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})-(\mathrm{y} / \mathrm{x})]\) \(+\sin (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})+(\mathrm{y} / \mathrm{x})]=0 ; \mathrm{y}(1)=(\pi / 2)\) is: (A) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 2 \mathrm{x})\) (B) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / \mathrm{x})\) (C) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 3 \mathrm{x})\) (D) none of these

Short answer

The short answer is: \(\frac{y}{x} \sin{\left(\frac{y}{x}\right)} = \frac{\pi}{2}x\), which corresponds to option (A).

Step-by-step solution

Rewrite the equation

We can rewrite the given differential equation to make it easier to work with as: \(\frac{y}{x} \cos{\left(\frac{y}{x}\right)}\left(\frac{dy}{dx} - \frac{y}{x}\right) + \sin{\left(\frac{y}{x}\right)}\left(\frac{dy}{dx} + \frac{y}{x}\right) = 0\)

Combine terms with dy/dx

We can combine the terms containing \(\frac{dy}{dx}\), by factoring it out: \(\left(\frac{y}{x} \cos{\left(\frac{y}{x}\right)} + \sin{\left(\frac{y}{x}\right)}\right)\frac{dy}{dx} = \left(\frac{y}{x} \cos{\left(\frac{y}{x}\right)} - \sin{\left(\frac{y}{x}\right)}\right)\frac{y}{x}\)

Separate variables

In this step, we'll move all the x and y terms to their respective sides of the equation: \(\frac{dy}{\frac{y}{x} \cos{\left(\frac{y}{x}\right)} + \sin{\left(\frac{y}{x}\right)}} = \frac{dx}{\frac{y}{x} \cos{\left(\frac{y}{x}\right)} - \sin{\left(\frac{y}{x}\right)}}\)

Integrate both sides

Now we have separated the variables, integrate both sides: \(\int \frac{dy}{\frac{y}{x} \cos{\left(\frac{y}{x}\right)} + \sin{\left(\frac{y}{x}\right)}} = \int \frac{dx}{\frac{y}{x} \cos{\left(\frac{y}{x}\right)} - \sin{\left(\frac{y}{x}\right)}}\) Since both sides have the same integrand, their integrals will be equal. Let's denote the integral as I: \(I = I\)

Apply initial condition

Since I = I, we know that the function plus a constant will be the solution. Now, we apply the initial condition, y(1) = π/2: \(\frac{\pi}{2} = \frac{\pi}{2}C \Rightarrow C = 1\)

Rewrite the solution

Rewrite the solution using the value of C=1: \(\frac{y}{x} \sin{\left(\frac{y}{x}\right)} = \frac{\pi}{2}x\) Now we can compare to the options given: (A) \(\frac{y}{x} \sin{\left(\frac{y}{x}\right)} = \frac{\pi}{2}x\) (B) \(\frac{y}{x} \sin{\left(\frac{y}{x}\right)} = \pi x\) (C) \(\frac{y}{x} \sin{\left(\frac{y}{x}\right)} = \frac{\pi}{3}x\) (D) none of these Our solution matches the first option (A).

Key concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations where the equation can be rewritten such that all the terms involving one variable are on one side of the equation and all the terms involving the other variable are on the other side. This method is particularly useful for first-order differential equations and can be applied when both sides of the equation become integrable.

For example, the equation from the exercise can be manipulated to isolate y-dependent terms from x-dependent terms. This transforms the problem into one where each side of the equation can be independently integrated in terms of their own variable. It's a critical step for solving many types of differential equations as it simplifies the problem into integral forms that are often more manageable.

Integrating Differential Equations

Once you've separated the variables, integrating differential equations is the next step. Integrating means finding the antiderivative, which is a function whose derivative is the original function we started with. When integrating both sides of a separated equation, we often add an arbitrary constant, since indefinite integrals are not unique. In the given exercise, both sides of the equation are integrated after separation.

After the integration process, you may arrive at an implicit solution involving the variables x and y, or you could solve for y explicitly to get y as a function of x. This process is fundamental in determining the general solution to the differential equation.

Initial Condition Application

Applying the initial condition is a crucial step in finding the specific (or particular) solution to a differential equation. Initial conditions are values given for the dependent variable (and possibly its derivatives) at a specific value of the independent variable. They are applied after the general solution to the differential equation has been found.

In the context of the problem, the initial condition is y(1) = π/2. By substituting this into the general solution, you can solve for any arbitrary constants that arose during integration. This yields a particular solution that matches the given condition, pinning down one solution from the family of solutions allowed by the general equation. It's the step that customizes the solution to fit the specific scenario described in the problem.