Question

Solve the differential equation \(\frac{x d x+y d y}{\sqrt{x^{2}+y^{2}}}=\frac{y d x-x d y}{x^{2}}\)

Short answer

Based on the provided step-by-step solution to the differential equation \(\frac{x dx + y dy}{\sqrt{x^2+y^2}} = \frac{y dx - x dy}{x^2}\), identify the final general solution. The general solution for the given differential equation is \(y_1 = \sqrt{\frac{-x + \sqrt{x^2 - 8 (x^2-1)}}{4}}\) and \(y_2 = -\sqrt{\frac{-x + \sqrt{x^2 - 8 (x^2-1)}}{4}}\).

Step-by-step solution

Identify total derivatives

Let's start by rewriting the given equation in a more familiar differential equation form. First, multiply both sides of the equation by the denominators of the fractions: \((x dx + y dy)(x^2) = (\sqrt{x^2 + y^2})(y dx - x dy)\) Now, we have: \(x^3 dx + x^2 y dy = y \sqrt{x^2 + y^2} dx - x \sqrt{x^2 + y^2} dy\) Now, it's crucial to identify the total derivatives. Observe that \(\frac{dx}{dt} = x^2 + y^2\) and \(\frac{dy}{dt} = y - x\). To eliminate \(t\), we can differentiate \(\frac{dy}{dt}\) with respect to \(x\): \(\frac{d^2y}{dx^2}= \frac{d(y - x)}{dx}\) \(= \frac{dy}{dx} - 1\) Now we can rewrite the equation with these total derivatives and see if the equation simplifies.

Make substitutions and simplify

Replace \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) with their respective expressions, and the given differential equation becomes: \(x^3\frac{dx}{dt} + x^2 y \frac{dy}{dt} = y \sqrt{\left(\frac{dx}{dt}\right)^2} - x \sqrt{\left(\frac{dy}{dt} - 1\right)^2}\) Substitute the identified total derivatives: \(x^3(x^2 + y^2) + x^2 y(y-x) = y \sqrt{(x^2+y^2)^2} - x \sqrt{(y-x)^2}\) Now, simplify the left side of the equation: \(x^5 + x^3 y^2 + x^3 y^2 - x^4 y = y (x^2+y^2) - x (y-x)\) Combine the terms: \(x^5 + 2x^3 y^2 - x^4 y = x^3 y - x^4\) Now, we can further simplify by dividing both sides of the equation by \(x^3\): \(x^2 + 2y^2 - xy = y - x\)

Solve the simplified equation for y

Now, the equation is simplified to a form that can be solved for \(y\): \(y - x = x^2 + 2y^2 - xy\) Rearrange the terms: \(0 = x^2 + x - 2y^2 + xy\) This is a quadratic equation in terms of \(y^2\). Solve for \(y^2\) by using the quadratic formula: \( y^2 = \frac{-x \pm \sqrt{x^2 - 4(2)(x^2-1)}}{4}\) Now, take the square root on both sides: \(y = \pm \sqrt{\frac{-x \pm \sqrt{x^2 - 8 (x^2-1)}}{4}}\) This gives us the general solution for the given differential equation in the form of two solutions: \(y_1 = \sqrt{\frac{-x + \sqrt{x^2 - 8 (x^2-1)}}{4}}\) and \(y_2 = -\sqrt{\frac{-x + \sqrt{x^2 - 8 (x^2-1)}}{4}}\)

Key concepts

Understanding Total Derivatives

Total derivatives play a pivotal role in solving differential equations, especially when multiple variables are involved. In the context of calculus, the total derivative of a function reflects the rate of change of that function with respect to all of its variables.

For example, if we have a function that depends on two variables, say time and position, the total derivative would account for how both these variables affect the function's rate of change simultaneously. In the exercise provided, the recognition of total derivatives is the first critical step towards simplifying and eventually solving the differential equation.

Understanding and identifying total derivatives allow us to express complex relationships and changes in a more manageable form. They are the essential building blocks that lead us to a point where we can use other techniques, such as separation of variables or integrating factors, to find the solution to a differential equation. This transformation is crucial for progressing towards a solution that is not immediately obvious by just looking at the original equation.

Solving Quadratic Equations

Quadratic equations, which are polynomials of degree two, often arise in various areas of mathematics and science. They are typically in the form of \( ax^2 + bx + c = 0 \), and solving them involves finding the values of \( x \) that make the equation true.

The general solution to a quadratic equation can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This is exactly what is applied in the exercise when we encounter the equation in the simplified form.

The quadratic formula often yields two solutions, representing the points where the quadratic curve (a parabola) intersects the x-axis. In the differential equation presented in the exercise, the quadratic formula is adapted to solve for \( y^2 \), which is a crucial step in isolating the dependent variable \( y \) and finding the general solution to the equation.

Integral Calculus in Differential Equations

Integral calculus is the study of integrals and their properties. It is commonly used together with differential calculus because they are inverse processes of each other. While differential calculus is concerned with the concept of a derivative, integral calculus focuses on the concept of an integral as the area under a curve.

However, in the context of solving differential equations, integral calculus is not always directly applied. Instead, it often underlies the methods used for solving these equations. For instance, after finding the total derivative of a function, we may need to integrate it to find the original function. In the solution process for our exercise, once the differential equation has been segregated into a familiar form using total derivatives, we could potentially use integral calculus to find a function that agrees with this derivative.

In many complex differential equations, integration may not happen explicitly, but the methods employed may have their roots in integral calculus. Integral calculus serves as the foundational concept that allows us to reverse the process of differentiation, thus revealing the functions that are hidden within the derivatives.