Question

Solve \(\left(\frac{\sin 2 x}{y}+x\right) d x+\left(y-\frac{\sin ^{2} x}{y^{2}}\right) d y=0\).

Short answer

After attempting different methods and simplifications, we were unable to find a closed-form solution to the given equation. The simplified form involving the substitution \(v = \frac{\sin{x}}{y}\) led to a first-order linear differential equation, but with an integral involving an integrating factor that could not be solved in closed form. Therefore, we conclude that the initial equation was already in the simplest form we could reach and cannot be solved further in closed form.

Step-by-step solution

Separate variables

Let's rewrite the given equation to achieve the form \(\frac{dy}{dx} = F(x, y)\). $$\frac{\sin{2x}}{y}dx + xdx + ydy = \frac{\sin^{2}{x}}{y^2}dy$$ Now, let's get all terms involving x and dx on one side and all terms involving y and dy on the other side. $$\left(\frac{\sin{2x}}{y} + x\right) dx = \left(y - \frac{\sin^{2}{x}}{y^{2}}\right) dy$$ Finally, we can rewrite the equation as: $$\frac{dy}{dx} = \frac{\frac{\sin{2x}}{y} + x}{y - \frac{\sin^{2}{x}}{y^{2}}}$$

Identify the type of differential equation

The equation has both variables x and y on both sides, and it doesn't fall into any familiar categories like separable, exact, or linear. So, we don't have any immediate method to solve it. However, we can notice that the numerator and denominator on the right side are both polynomials in x and y, so it may be possible to simplify the equation and find a suitable substitution.

Simplification and substitution

Let's try to find a substitution to simplify the equation. Let's denote \(v = \frac{\sin{x}}{y}\), then it follows that: $$yv = \sin{x} \Rightarrow y = \frac{\sin{x}}{v} \Rightarrow dy = \frac{v\cos{x} dx - \sin{x} dv}{v^2}$$ Replacing in the original equation we get: $$\left(\frac{2v + x}{1 - v^2}\right) dx = \left(\frac{\sin^2{x}}{y^2} - y\right) \left(\frac{v\cos{x} dx - \sin{x} dv}{v^2}\right)$$ Now we can replace y back into the equation, and simplify $$\left(\frac{2v + x}{1 - v^2}\right) dx = \left(\frac{\sin^2{x}}{\sin^2{x}/v^2} - \frac{\sin{x}}{v}\right) \left(\frac{v\cos{x} dx - \sin{x} dv}{v^2}\right)$$ This simplifies to: $$\left(\frac{2v + x}{1 - v^2}\right) dx = \left(v^2 - v\right) \left(\frac{v\cos{x} dx - \sin{x} dv}{v^2}\right)$$ Canceling out a factor of v on the right, we have: $$\left(\frac{2v + x}{1 - v^2}\right) dx = (v - 1)\left(v\cos{x} dx - \sin{x} dv\right)$$ Now, we have a first-order linear differential equation in v and x.

Solve the first-order linear differential equation

The resulting equation is a linear first-order differential equation in v and x. It can be solved using an integrating factor.})(); The equation can be written in the form $$\frac{dv}{dx} + Pv = Q$$ From the equation we have, $$P = -\frac{2v + x}{1 - v^2}, Q = -\frac{\sin{x}}{1 - v^2}\cos{x}$$ Find the integrating factor, $$\text{IF} = e^{\int P dx} = e^{-\int \frac{2v + x}{1 - v^2}dx}$$ This is an integral we cannot solve in closed form, so we cannot proceed any further. The initial equation was already in the simplest form that we could reach.  We had the following equation $$\frac{dy}{dx} = \frac{\frac{\sin{2x}}{y} + x}{y - \frac{\sin^{2}{x}}{y^{2}}}$$

Key concepts

Integral Calculus

Integral calculus is one of the two principal branches of calculus, with the other being differential calculus. It focuses on the accumulation of quantities and the calculation of areas under curves. In the context of differential equations, integral calculus plays a critical role in finding solutions. Integral calculus allows us to reverse the process of differentiation, which is essential when we work on solving a differential equation since the solution often involves finding an antiderivative or integrating a function.

For instance, when solving a first-order linear differential equation, we may use an integrating factor, which involves the exponentiation of the integral of a given function. This approach utilizes the methods of integral calculus to simplify the equation into a form where the solution becomes more apparent or attainable.

IIT JEE Main and Advanced Mathematics

The IIT JEE Main and Advanced examinations are highly competitive entrance tests for engineering aspirants in India, aiming to secure a place in prestigious Indian Institutes of Technology (IITs) and other top engineering colleges. Mathematics is a core subject of these exams and includes topics from algebra, trigonometry, calculus, and geometry.

Differential equations, which appear in the ‘Calculus’ section, are particularly important for these exams. To excel in IIT JEE, students must be adept at solving complex differential equations, as they are frequently included in the advanced mathematics section of the test. The exercises and problems are designed to challenge a student's understanding of mathematical concepts, application skills, and problem-solving abilities. Successful candidates typically have a strong grasp of integral calculus and the techniques required for solving different types of differential equations, including first-order linear differential equations.

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation where the derivative of the unknown function appears to the first power and no higher. These equations take the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). Solving these often involves finding an integrating factor, which is a function that, when multiplied by the differential equation, makes the left side of the equation directly integrable.

The process of identifying and solving these equations is fundamental in many fields, including physics, engineering, and economics. For the exercise given, the challenge lies in transforming the complex-looking equation into a standard first-order linear differential equation or finding a suitable substitution that simplifies the equation. The integrating factor method is crucial for progress, even though, as the provided steps indicate, sometimes finding an integrating factor in closed form is not possible, leaving the equation in its simplest form unreachable by standard methods.